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<h1 class="title d-none d-lg-block"><span class="chapter-number">41</span>&nbsp; <span class="chapter-title">Improper Integrals</span></h1>
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<p>This section uses these add-on packages:</p>
<div class="sourceCode cell-code" id="cb1"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb1-1"><a href="#cb1-1" aria-hidden="true" tabindex="-1"></a><span class="im">using</span> <span class="bu">CalculusWithJulia</span></span>
<span id="cb1-2"><a href="#cb1-2" aria-hidden="true" tabindex="-1"></a><span class="im">using</span> <span class="bu">Plots</span></span>
<span id="cb1-3"><a href="#cb1-3" aria-hidden="true" tabindex="-1"></a><span class="im">using</span> <span class="bu">SymPy</span></span>
<span id="cb1-4"><a href="#cb1-4" aria-hidden="true" tabindex="-1"></a><span class="im">using</span> <span class="bu">QuadGK</span></span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<hr>
<p>A function <span class="math inline">\(f(x)\)</span> is Riemann integrable over an interval <span class="math inline">\([a,b]\)</span> if some limit involving Riemann sums exists. This limit will fail to exist if <span class="math inline">\(f(x) = \infty\)</span> in <span class="math inline">\([a,b]\)</span>. As well, the Riemann sum idea is undefined if either <span class="math inline">\(a\)</span> or <span class="math inline">\(b\)</span> (or both) are infinite, so the limit won’t exist in this case.</p>
<p>To define integrals with either functions having singularities or infinite domains, the idea of an improper integral is introduced with definitions to handle the two cases above.</p>
<div class="cell" data-cache="true" data-hold="true" data-execution_count="4">
<div class="cell-output cell-output-display" data-execution_count="5">
<div class="d-flex justify-content-center"> <figure class="figure"> <img src="data:image/gif;base64,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
<figcaption class="figure-caption"><div class="markdown"><p>Area under \(1/\sqrt{x}\) over \([a,b]\) increases as \(a\) gets closer to \(0\). Will it grow unbounded or have a limit?</p>
</div> </figcaption>
</figure>
</div>
</div>
</div>
<section id="infinite-domains" class="level2" data-number="41.1">
<h2 data-number="41.1" class="anchored" data-anchor-id="infinite-domains"><span class="header-section-number">41.1</span> Infinite domains</h2>
<p>Let <span class="math inline">\(f(x)\)</span> be a reasonable function, so reasonable that for any <span class="math inline">\(a &lt; b\)</span> the function is Riemann integrable, meaning <span class="math inline">\(\int_a^b f(x)dx\)</span> exists.</p>
<p>What needs to be the case so that we can discuss the integral over the entire real number line?</p>
<p>Clearly something. The function <span class="math inline">\(f(x) = 1\)</span> is reasonable by the idea above. Clearly the integral over and <span class="math inline">\([a,b]\)</span> is just <span class="math inline">\(b-a\)</span>, but the limit over an unbounded domain would be <span class="math inline">\(\infty\)</span>. Even though limits of infinity can be of interest in some cases, not so here. What will ensure that the area is finite over an infinite region?</p>
<p>Or is that even the right question. Now consider <span class="math inline">\(f(x) = \sin(\pi x)\)</span>. Over every interval of the type <span class="math inline">\([-2n, 2n]\)</span> the area is <span class="math inline">\(0\)</span>, and over any interval, <span class="math inline">\([a,b]\)</span> the area never gets bigger than <span class="math inline">\(2\)</span>. But still this function does not have a well defined area on an infinite domain.</p>
<p>The right question involves a limit. Fix a finite <span class="math inline">\(a\)</span>. We define the definite integral over <span class="math inline">\([a,\infty)\)</span> to be</p>
<p><span class="math display">\[
\int_a^\infty f(x) dx = \lim_{M \rightarrow \infty} \int_a^M f(x) dx,
\]</span></p>
<p>when the limit exists. Similarly, we define the definite integral over <span class="math inline">\((-\infty, a]\)</span> through</p>
<p><span class="math display">\[
\int_{-\infty}^a f(x) dx = \lim_{M \rightarrow -\infty} \int_M^a f(x) dx.
\]</span></p>
<p>For the interval <span class="math inline">\((-\infty, \infty)\)</span> we have need <em>both</em> these limits to exist, and then:</p>
<p><span class="math display">\[
\int_{-\infty}^\infty f(x) dx = \lim_{M \rightarrow -\infty} \int_M^a f(x) dx + \lim_{M \rightarrow \infty} \int_a^M f(x) dx.
\]</span></p>
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<div class="callout-caption-container flex-fill">
Note
</div>
</div>
<div class="callout-body-container callout-body">
<p>When the integral exists, it is said to <em>converge</em>. If it doesn’t exist, it is said to <em>diverge</em>.</p>
</div>
</div>
<section id="examples" class="level5">
<h5 class="anchored" data-anchor-id="examples">Examples</h5>
<ul>
<li>The function <span class="math inline">\(f(x) = 1/x^2\)</span> is integrable over <span class="math inline">\([1, \infty)\)</span>, as this limit exists:</li>
</ul>
<p><span class="math display">\[
\lim_{M \rightarrow \infty} \int_1^M \frac{1}{x^2}dx = \lim_{M \rightarrow \infty} -\frac{1}{x}\big|_1^M
= \lim_{M \rightarrow \infty} 1 - \frac{1}{M} = 1.
\]</span></p>
<ul>
<li>The function <span class="math inline">\(f(x) = 1/x^{1/2}\)</span> is not integrable over <span class="math inline">\([1, \infty)\)</span>, as this limit fails to exist:</li>
</ul>
<p><span class="math display">\[
\lim_{M \rightarrow \infty} \int_1^M \frac{1}{x^{1/2}}dx = \lim_{M \rightarrow \infty} \frac{x^{1/2}}{1/2}\big|_1^M
= \lim_{M \rightarrow \infty} 2\sqrt{M} - 2 = \infty.
\]</span></p>
<p>The limit is infinite, so does not exist except in an extended sense.</p>
<ul>
<li>The function <span class="math inline">\(x^n e^{-x}\)</span> for <span class="math inline">\(n = 1, 2, \dots\)</span> is integrable over <span class="math inline">\([0,\infty)\)</span>.</li>
</ul>
<p>Before showing this, we recall the fundamental theorem of calculus. The limit existing is the same as saying the limit of <span class="math inline">\(F(M) - F(a)\)</span> exists for an antiderivative of <span class="math inline">\(f(x)\)</span>.</p>
<p>For this particular problem, it can be shown by integration by parts that for positive, integer values of <span class="math inline">\(n\)</span> that an antiderivative exists of the form <span class="math inline">\(F(x) = p(x)e^{-x}\)</span>, where <span class="math inline">\(p(x)\)</span> is a polynomial of degree <span class="math inline">\(n\)</span>. But we’ve seen that for any <span class="math inline">\(n&gt;0\)</span>, <span class="math inline">\(\lim_{x \rightarrow \infty} x^n e^{-x} = 0\)</span>, so the same is true for any polynomial. So, <span class="math inline">\(\lim_{M \rightarrow \infty} F(M) - F(1) = -F(1)\)</span>.</p>
<ul>
<li>The function <span class="math inline">\(e^x\)</span> is integrable over <span class="math inline">\((-\infty, a]\)</span> but not</li>
</ul>
<p><span class="math display">\[
[a, \infty)
\]</span></p>
<p>for any finite <span class="math inline">\(a\)</span>. This is because, <span class="math inline">\(F(M) = e^x\)</span> and this has a limit as <span class="math inline">\(x\)</span> goes to <span class="math inline">\(-\infty\)</span>, but not <span class="math inline">\(\infty\)</span>.</p>
<ul>
<li><p>Let <span class="math inline">\(f(x) = x e^{-x^2}\)</span>. This function has an integral over <span class="math inline">\([0, \infty)\)</span> and more generally <span class="math inline">\((-\infty, \infty)\)</span>. To see, we note that as it is an odd function, the area from <span class="math inline">\(0\)</span> to <span class="math inline">\(M\)</span> is the opposite sign of that from <span class="math inline">\(-M\)</span> to <span class="math inline">\(0\)</span>. So <span class="math inline">\(\lim_{M \rightarrow \infty} (F(M) - F(0)) = \lim_{M \rightarrow -\infty} (F(0) - (-F(\lvert M\lvert)))\)</span>. We only then need to investigate the one limit. But we can see by substitution with <span class="math inline">\(u=x^2\)</span>, that an antiderivative is <span class="math inline">\(F(x) = (-1/2) \cdot e^{-x^2}\)</span>. Clearly, <span class="math inline">\(\lim_{M \rightarrow \infty}F(M) = 0\)</span>, so the answer is well defined, and the area from <span class="math inline">\(0\)</span> to <span class="math inline">\(\infty\)</span> is just <span class="math inline">\(e/2\)</span>. From <span class="math inline">\(-\infty\)</span> to <span class="math inline">\(0\)</span> it is <span class="math inline">\(-e/2\)</span> and the total area is <span class="math inline">\(0\)</span>, as the two sides “cancel” out.</p></li>
<li><p>Let <span class="math inline">\(f(x) = \sin(x)\)</span>. Even though <span class="math inline">\(\lim_{M \rightarrow \infty} (F(M) - F(-M) ) = 0\)</span>, this function is not integrable. The fact is we need <em>both</em> the limit <span class="math inline">\(F(M)\)</span> and <span class="math inline">\(F(-M)\)</span> to exist as <span class="math inline">\(M\)</span> goes to <span class="math inline">\(\infty\)</span>. In this case, even though the area cancels if <span class="math inline">\(\infty\)</span> is approached at the same rate, this isn’t sufficient to guarantee the two limits exists independently.</p></li>
<li><p>Will the function <span class="math inline">\(f(x) = 1/(x\cdot(\log(x))^2)\)</span> have an integral over <span class="math inline">\([e, \infty)\)</span>?</p></li>
</ul>
<p>We first find an antiderivative using the <span class="math inline">\(u\)</span>-substitution <span class="math inline">\(u(x) = \log(x)\)</span>:</p>
<p><span class="math display">\[
\int_e^M \frac{e}{x \log(x)^{2}} dx
= \int_{\log(e)}^{\log(M)} \frac{1}{u^{2}} du
= \frac{-1}{u} \big|_{1}^{\log(M)}
= \frac{-1}{\log(M)} - \frac{-1}{1}
= 1 - \frac{1}{M}.
\]</span></p>
<p>As <span class="math inline">\(M\)</span> goes to <span class="math inline">\(\infty\)</span>, this will converge to <span class="math inline">\(1\)</span>.</p>
<ul>
<li>The sinc function <span class="math inline">\(f(x) = \sin(\pi x)/(\pi x)\)</span> does not have a nice antiderivative. Seeing if the limit exists is a bit of a problem. However, this function is important enough that there is a built-in function, <code>Si</code>, that computes <span class="math inline">\(\int_0^x \sin(u)/u\cdot du\)</span>. This function can be used through <code>sympy.Si(...)</code>:</li>
</ul>
<div class="cell" data-execution_count="5">
<div class="sourceCode cell-code" id="cb2"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb2-1"><a href="#cb2-1" aria-hidden="true" tabindex="-1"></a><span class="pp">@syms</span> M</span>
<span id="cb2-2"><a href="#cb2-2" aria-hidden="true" tabindex="-1"></a><span class="fu">limit</span>(sympy.<span class="fu">Si</span>(M), M <span class="op">=&gt;</span> oo)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="6">
<span class="math-left-align" style="padding-left: 4px; width:0; float:left;">
\[
\frac{\pi}{2}
\]
</span>
</div>
</div>
</section>
<section id="numeric-integration" class="level3" data-number="41.1.1">
<h3 data-number="41.1.1" class="anchored" data-anchor-id="numeric-integration"><span class="header-section-number">41.1.1</span> Numeric integration</h3>
<p>The <code>quadgk</code> function (available through <code>QuadGK</code>) is able to accept <code>Inf</code> and <code>-Inf</code> as endpoints of the interval. For example, this will integrate <span class="math inline">\(e^{-x^2/2}\)</span> over the real line:</p>
<div class="cell" data-execution_count="6">
<div class="sourceCode cell-code" id="cb3"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb3-1"><a href="#cb3-1" aria-hidden="true" tabindex="-1"></a><span class="fu">f</span>(x) <span class="op">=</span> <span class="fu">exp</span>(<span class="op">-</span>x<span class="op">^</span><span class="fl">2</span><span class="op">/</span><span class="fl">2</span>)</span>
<span id="cb3-2"><a href="#cb3-2" aria-hidden="true" tabindex="-1"></a><span class="fu">quadgk</span>(f, <span class="op">-</span><span class="cn">Inf</span>, <span class="cn">Inf</span>)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="7">
<pre><code>(2.506628274639168, 3.608438072243189e-8)</code></pre>
</div>
</div>
<p>(If may not be obvious, but this is <span class="math inline">\(\sqrt{2\pi}\)</span>.)</p>
</section>
</section>
<section id="singularities" class="level2" data-number="41.2">
<h2 data-number="41.2" class="anchored" data-anchor-id="singularities"><span class="header-section-number">41.2</span> Singularities</h2>
<p>Suppose <span class="math inline">\(\lim_{x \rightarrow c}f(x) = \infty\)</span> or <span class="math inline">\(-\infty\)</span>. Then a Riemann sum that contains an interval including <span class="math inline">\(c\)</span> will not be finite if the point chosen in the interval is <span class="math inline">\(c\)</span>. Though we could choose another point, this is not enough as the definition must hold for any choice of the <span class="math inline">\(c_i\)</span>.</p>
<p>However, if <span class="math inline">\(c\)</span> is isolated, we can get close to <span class="math inline">\(c\)</span> and see how the area changes.</p>
<p>Suppose <span class="math inline">\(a &lt; c\)</span>, we define <span class="math inline">\(\int_a^c f(x) dx = \lim_{M \rightarrow c-} \int_a^c f(x) dx\)</span>. If this limit exists, the definite integral with <span class="math inline">\(c\)</span> is well defined. Similarly, the integral from <span class="math inline">\(c\)</span> to <span class="math inline">\(b\)</span>, where <span class="math inline">\(b &gt; c\)</span>, can be defined by a right limit going to <span class="math inline">\(c\)</span>. The integral from <span class="math inline">\(a\)</span> to <span class="math inline">\(b\)</span> will exist if both the limits are finite.</p>
<section id="examples-1" class="level5">
<h5 class="anchored" data-anchor-id="examples-1">Examples</h5>
<ul>
<li>Consider the example of the initial illustration, <span class="math inline">\(f(x) = 1/\sqrt{x}\)</span> at <span class="math inline">\(0\)</span>. Here <span class="math inline">\(f(0)= \infty\)</span>, so the usual notion of a limit won’t apply to <span class="math inline">\(\int_0^1 f(x) dx\)</span>. However,</li>
</ul>
<p><span class="math display">\[
\lim_{M \rightarrow 0+} \int_M^1 \frac{1}{\sqrt{x}} dx
= \lim_{M \rightarrow 0+} \frac{\sqrt{x}}{1/2} \big|_M^1
= \lim_{M \rightarrow 0+} 2(1) - 2\sqrt{M} = 2.
\]</span></p>
<div class="callout-note callout callout-style-default callout-captioned">
<div class="callout-header d-flex align-content-center">
<div class="callout-icon-container">
<i class="callout-icon"></i>
</div>
<div class="callout-caption-container flex-fill">
Note
</div>
</div>
<div class="callout-body-container callout-body">
<p>The cases <span class="math inline">\(f(x) = x^{-n}\)</span> for <span class="math inline">\(n &gt; 0\)</span> are tricky to keep straight. For <span class="math inline">\(n &gt; 1\)</span>, the functions can be integrated over <span class="math inline">\([1,\infty)\)</span>, but not <span class="math inline">\((0,1]\)</span>. For <span class="math inline">\(0 &lt; n &lt; 1\)</span>, the functions can be integrated over <span class="math inline">\((0,1]\)</span> but not <span class="math inline">\([1, \infty)\)</span>.</p>
</div>
</div>
<ul>
<li>Now consider <span class="math inline">\(f(x) = 1/x\)</span>. Is this integral <span class="math inline">\(\int_0^1 1/x \cdot dx\)</span> defined? It will be <em>if</em> this limit exists:</li>
</ul>
<p><span class="math display">\[
\lim_{M \rightarrow 0+} \int_M^1 \frac{1}{x} dx
= \lim_{M \rightarrow 0+} \log(x) \big|_M^1
= \lim_{M \rightarrow 0+} \log(1) - \log(M) = \infty.
\]</span></p>
<p>As the limit does not exist, the function is not integrable around <span class="math inline">\(0\)</span>.</p>
<ul>
<li><code>SymPy</code> may give answers which do not coincide with our definitions, as it uses complex numbers as a default assumption. In this case it returns the proper answer when integrated from <span class="math inline">\(0\)</span> to <span class="math inline">\(1\)</span> and <code>NaN</code> for an integral over <span class="math inline">\((-1,1)\)</span>:</li>
</ul>
<div class="cell" data-execution_count="7">
<div class="sourceCode cell-code" id="cb5"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb5-1"><a href="#cb5-1" aria-hidden="true" tabindex="-1"></a><span class="pp">@syms</span> x</span>
<span id="cb5-2"><a href="#cb5-2" aria-hidden="true" tabindex="-1"></a><span class="fu">integrate</span>(<span class="fl">1</span><span class="op">/</span>x, (x, <span class="fl">0</span>, <span class="fl">1</span>)), <span class="fu">integrate</span>(<span class="fl">1</span><span class="op">/</span>x, (x, <span class="op">-</span><span class="fl">1</span>, <span class="fl">1</span>))</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="8">
<pre><code>(oo, nan)</code></pre>
</div>
</div>
<ul>
<li>Suppose you know <span class="math inline">\(\int_1^\infty x^2 f(x) dx\)</span> exists. Does this imply <span class="math inline">\(\int_0^1 f(1/x) dx\)</span> exists?</li>
</ul>
<p>We need to consider the limit of <span class="math inline">\(\int_M^1 f(1/x) dx\)</span>. We try the <span class="math inline">\(u\)</span>-substitution <span class="math inline">\(u(x) = 1/x\)</span>. This gives <span class="math inline">\(du = -(1/x^2)dx = -u^2 dx\)</span>. So, the substitution becomes:</p>
<p><span class="math display">\[
\int_M^1 f(1/x) dx = \int_{1/M}^{1/1} f(u) (-u^2) du = \int_1^{1/M} u^2 f(u) du.
\]</span></p>
<p>But the limit as <span class="math inline">\(M \rightarrow 0\)</span> of <span class="math inline">\(1/M\)</span> is the same going to <span class="math inline">\(\infty\)</span>, so the right side will converge by the assumption. Thus we get <span class="math inline">\(f(1/x)\)</span> is integrable over <span class="math inline">\((0,1]\)</span>.</p>
</section>
<section id="numeric-integration-1" class="level3" data-number="41.2.1">
<h3 data-number="41.2.1" class="anchored" data-anchor-id="numeric-integration-1"><span class="header-section-number">41.2.1</span> Numeric integration</h3>
<p>So far our use of the <code>quadgk</code> function specified the region to integrate via <code>a</code>, <code>b</code>, as in <code>quadgk(f, a, b)</code>. In fact, it can specify values in between for which the function should not be sampled. For example, were we to integrate <span class="math inline">\(1/\sqrt{\lvert x\rvert}\)</span> over <span class="math inline">\([-1,1]\)</span>, we would want to avoid <span class="math inline">\(0\)</span> as a point to sample. Here is how:</p>
<div class="cell" data-hold="true" data-execution_count="8">
<div class="sourceCode cell-code" id="cb7"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb7-1"><a href="#cb7-1" aria-hidden="true" tabindex="-1"></a><span class="fu">f</span>(x) <span class="op">=</span> <span class="fl">1</span> <span class="op">/</span> <span class="fu">sqrt</span>(<span class="fu">abs</span>(x))</span>
<span id="cb7-2"><a href="#cb7-2" aria-hidden="true" tabindex="-1"></a><span class="fu">quadgk</span>(f, <span class="op">-</span><span class="fl">1</span>, <span class="fl">0</span>, <span class="fl">1</span>)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="9">
<pre><code>(3.999999962817228, 5.736423067171012e-8)</code></pre>
</div>
</div>
<p>Just trying <code>quadgk(f, -1, 1)</code> leads to a <code>DomainError</code>, as <code>0</code> will be one of the points sampled. The general call is like <code>quadgk(f, a, b, c, d,...)</code> which integrates over <span class="math inline">\((a,b)\)</span> and <span class="math inline">\((b,c)\)</span> and <span class="math inline">\((c,d)\)</span>, <span class="math inline">\(\dots\)</span>. The algorithm is not supposed to evaluate the function at the endpoints of the intervals.</p>
</section>
</section>
<section id="probability-applications" class="level2" data-number="41.3">
<h2 data-number="41.3" class="anchored" data-anchor-id="probability-applications"><span class="header-section-number">41.3</span> Probability applications</h2>
<p>A probability density is a function <span class="math inline">\(f(x) \geq 0\)</span> which is integrable on <span class="math inline">\((-\infty, \infty)\)</span> and for which <span class="math inline">\(\int_{-\infty}^\infty f(x) dx =1\)</span>. The cumulative distribution function is defined by <span class="math inline">\(F(x)=\int_{-\infty}^x f(u) du\)</span>.</p>
<p>Probability densities are good example of using improper integrals.</p>
<ul>
<li>Show that <span class="math inline">\(f(x) = (1/\pi) (1/(1 + x^2))\)</span> is a probability density function.</li>
</ul>
<p>We need to show that the integral exists and is <span class="math inline">\(1\)</span>. For this, we use the fact that <span class="math inline">\((1/\pi) \cdot \tan^{-1}(x)\)</span> is an antiderivative. Then we have:</p>
<p><span class="math display">\[
\lim_{M \rightarrow \infty} F(M) = (1/\pi) \cdot \pi/2
\]</span></p>
<p>and as <span class="math inline">\(\tan^{-1}(x)\)</span> is odd, we must have <span class="math inline">\(F(-\infty) = \lim_{M \rightarrow -\infty} f(M) = -(1/\pi) \cdot \pi/2\)</span>. All told, <span class="math inline">\(F(\infty) - F(-\infty) = 1/2 - (-1/2) = 1\)</span>.</p>
<ul>
<li>Show that <span class="math inline">\(f(x) = 1/(b-a)\)</span> for <span class="math inline">\(a \leq x \leq b\)</span> and <span class="math inline">\(0\)</span> otherwise is a probability density.</li>
</ul>
<p>The integral for <span class="math inline">\(-\infty\)</span> to <span class="math inline">\(a\)</span> of <span class="math inline">\(f(x)\)</span> is just an integral of the constant <span class="math inline">\(0\)</span>, so will be <span class="math inline">\(0\)</span>. (This is the only constant with finite area over an infinite domain.) Similarly, the integral from <span class="math inline">\(b\)</span> to <span class="math inline">\(\infty\)</span> will be <span class="math inline">\(0\)</span>. This means:</p>
<p><span class="math display">\[
\int_{-\infty}^\infty f(x) dx = \int_a^b \frac{1}{b-a} dx = 1.
\]</span></p>
<p>(One might also comment that <span class="math inline">\(f\)</span> is Riemann integrable on any <span class="math inline">\([0,M]\)</span> despite being discontinuous at <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span>.)</p>
<ul>
<li>Show that if <span class="math inline">\(f(x)\)</span> is a probability density then so is <span class="math inline">\(f(x-c)\)</span> for any <span class="math inline">\(c\)</span>.</li>
</ul>
<p>We have by the <span class="math inline">\(u\)</span>-substitution</p>
<p><span class="math display">\[
\int_{-\infty}^\infty f(x-c)dx = \int_{u(-\infty)}^{u(\infty)} f(u) du = \int_{-\infty}^\infty f(u) du = 1.
\]</span></p>
<p>The key is that we can use the regular <span class="math inline">\(u\)</span>-substitution formula provided <span class="math inline">\(\lim_{M \rightarrow \infty} u(M) = u(\infty)\)</span> is defined. (The <em>informal</em> notation <span class="math inline">\(u(\infty)\)</span> is defined by that limit.)</p>
<ul>
<li>If <span class="math inline">\(f(x)\)</span> is a probability density, then so is <span class="math inline">\((1/h) f((x-c)/h)\)</span> for any <span class="math inline">\(c, h &gt; 0\)</span>.</li>
</ul>
<p>Again, by a <span class="math inline">\(u\)</span> substitution with, now, <span class="math inline">\(u(x) = (x-c)/h\)</span>, we have <span class="math inline">\(du = (1/h) \cdot dx\)</span> and the result follows just as before:</p>
<p><span class="math display">\[
\int_{-\infty}^\infty \frac{1}{h}f(\frac{x-c}{h})dx = \int_{u(-\infty)}^{u(\infty)} f(u) du = \int_{-\infty}^\infty f(u) du = 1.
\]</span></p>
<ul>
<li>If <span class="math inline">\(F(x) = 1 - e^{-x}\)</span>, for <span class="math inline">\(x \geq 0\)</span>, and <span class="math inline">\(0\)</span> otherwise, find <span class="math inline">\(f(x)\)</span>.</li>
</ul>
<p>We want to just say <span class="math inline">\(F'(x)= e^{-x}\)</span> so <span class="math inline">\(f(x) = e^{-x}\)</span>. But some care is needed. First, that isn’t right. The derivative for <span class="math inline">\(x&lt;0\)</span> of <span class="math inline">\(F(x)\)</span> is <span class="math inline">\(0\)</span>, so <span class="math inline">\(f(x) = 0\)</span> if <span class="math inline">\(x &lt; 0\)</span>. What about for <span class="math inline">\(x&gt;0\)</span>? The derivative is <span class="math inline">\(e^{-x}\)</span>, but is that the right answer? <span class="math inline">\(F(x) = \int_{-\infty}^x f(u) du\)</span>, so we have to at least discuss if the <span class="math inline">\(-\infty\)</span> affects things. In this case, and in general the answer is <em>no</em>. For any <span class="math inline">\(x\)</span> we can find <span class="math inline">\(M &lt; x\)</span> so that we have <span class="math inline">\(F(x) = \int_{-\infty}^M f(u) du + \int_M^x f(u) du\)</span>. The first part is a constant, so will have derivative <span class="math inline">\(0\)</span>, the second will have derivative <span class="math inline">\(f(x)\)</span>, if the derivative exists (and it will exist at <span class="math inline">\(x\)</span> if the derivative is continuous in a neighborhood of <span class="math inline">\(x\)</span>).</p>
<p>Finally, at <span class="math inline">\(x=0\)</span> we have an issue, as <span class="math inline">\(F'(0)\)</span> does not exist. The left limit of the secant line approximation is <span class="math inline">\(0\)</span>, the right limit of the secant line approximation is <span class="math inline">\(1\)</span>. So, we can take <span class="math inline">\(f(x) = e^{-x}\)</span> for <span class="math inline">\(x &gt; 0\)</span> and <span class="math inline">\(0\)</span> otherwise, noting that redefining <span class="math inline">\(f(x)\)</span> at a point will not effect the integral as long as the point is finite.</p>
</section>
<section id="questions" class="level2" data-number="41.4">
<h2 data-number="41.4" class="anchored" data-anchor-id="questions"><span class="header-section-number">41.4</span> Questions</h2>
<section id="question" class="level6">
<h6 class="anchored" data-anchor-id="question">Question</h6>
<p>Is <span class="math inline">\(f(x) = 1/x^{100}\)</span> integrable around <span class="math inline">\(0\)</span>?</p>
<div class="cell" data-hold="true" data-execution_count="9">
<div class="cell-output cell-output-display" data-execution_count="10">
<form class="mx-2 my-3 mw-100" name="WeaveQuestion" data-id="13835909680318883433" data-controltype="">
<div class="form-group ">
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</section>
<section id="question-1" class="level6">
<h6 class="anchored" data-anchor-id="question-1">Question</h6>
<p>Is <span class="math inline">\(f(x) = 1/x^{1/3}\)</span> integrable around <span class="math inline">\(0\)</span>?</p>
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</section>
<section id="question-2" class="level6">
<h6 class="anchored" data-anchor-id="question-2">Question</h6>
<p>Is <span class="math inline">\(f(x) = x\cdot\log(x)\)</span> integrable on <span class="math inline">\([1,\infty)\)</span>?</p>
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</section>
<section id="question-3" class="level6">
<h6 class="anchored" data-anchor-id="question-3">Question</h6>
<p>Is <span class="math inline">\(f(x) = \log(x)/ x\)</span> integrable on <span class="math inline">\([1,\infty)\)</span>?</p>
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</section>
<section id="question-4" class="level6">
<h6 class="anchored" data-anchor-id="question-4">Question</h6>
<p>Is <span class="math inline">\(f(x) = \log(x)\)</span> integrable on <span class="math inline">\([1,\infty)\)</span>?</p>
<div class="cell" data-hold="true" data-execution_count="13">
<div class="cell-output cell-output-display" data-execution_count="14">
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No
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</section>
<section id="question-5" class="level6">
<h6 class="anchored" data-anchor-id="question-5">Question</h6>
<p>Compute the integral <span class="math inline">\(\int_0^\infty 1/(1+x^2) dx\)</span>.</p>
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</div>
</div>
</section>
<section id="question-6" class="level6">
<h6 class="anchored" data-anchor-id="question-6">Question</h6>
<p>Compute the the integral <span class="math inline">\(\int_1^\infty \log(x)/x^2 dx\)</span>.</p>
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</div>
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</div>
</div>
</section>
<section id="question-7" class="level6">
<h6 class="anchored" data-anchor-id="question-7">Question</h6>
<p>Compute the integral <span class="math inline">\(\int_0^2 (x-1)^{2/3} dx\)</span>.</p>
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<input id="14099481710229385942" type="number" class="form-control" placeholder="Numeric answer">
</div>
</div>
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</div>
</div>
</form>
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</div>
</div>
</section>
<section id="question-8" class="level6">
<h6 class="anchored" data-anchor-id="question-8">Question</h6>
<p>From the relationship that if <span class="math inline">\(0 \leq f(x) \leq g(x)\)</span> then <span class="math inline">\(\int_a^b f(x) dx \leq \int_a^b g(x) dx\)</span> it can be deduced that</p>
<ul>
<li>if <span class="math inline">\(\int_a^\infty f(x) dx\)</span> diverges, then so does <span class="math inline">\(\int_a^\infty g(x) dx\)</span>.</li>
<li>if <span class="math inline">\(\int_a^\infty g(x) dx\)</span> converges, then so does <span class="math inline">\(\int_a^\infty f(x) dx\)</span>.</li>
</ul>
<p>Let <span class="math inline">\(f(x) = \lvert \sin(x)/x^2 \rvert\)</span>.</p>
<p>What can you say about <span class="math inline">\(\int_1^\infty f(x) dx\)</span>, as <span class="math inline">\(f(x) \leq 1/x^2\)</span> on <span class="math inline">\([1, \infty)\)</span>?</p>
<div class="cell" data-hold="true" data-execution_count="17">
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It is convergent
</span>
</label>
</div>
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<label class="form-check-label" for="radio_2748116686296337231_2">
<input class="form-check-input" type="radio" name="radio_2748116686296337231" id="radio_2748116686296337231_2" value="2">
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It is divergent
</span>
</label>
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Can't say
</span>
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</div>
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<hr>
<p>Let <span class="math inline">\(f(x) = \lvert \sin(x) \rvert / x\)</span>.</p>
<p>What can you say about <span class="math inline">\(\int_1^\infty f(x) dx\)</span>, as <span class="math inline">\(f(x) \leq 1/x\)</span> on <span class="math inline">\([1, \infty)\)</span>?</p>
<div class="cell" data-hold="true" data-execution_count="18">
<div class="cell-output cell-output-display" data-execution_count="19">
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<input class="form-check-input" type="radio" name="radio_10586838793303528513" id="radio_10586838793303528513_1" value="1">
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It is convergent
</span>
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</div>
<div class="form-check">
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<input class="form-check-input" type="radio" name="radio_10586838793303528513" id="radio_10586838793303528513_2" value="2">
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It is divergent
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_10586838793303528513_3">
<input class="form-check-input" type="radio" name="radio_10586838793303528513" id="radio_10586838793303528513_3" value="3">
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Can't say
</span>
</label>
</div>
</div>
</div>
<div id="10586838793303528513_message" style="padding-bottom: 15px"></div>
</div>
</div>
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<hr>
<p>Let <span class="math inline">\(f(x) = 1/\sqrt{x^2 - 1}\)</span>. What can you say about <span class="math inline">\(\int_1^\infty f(x) dx\)</span>, as <span class="math inline">\(f(x) \geq 1/x\)</span> on <span class="math inline">\([1, \infty)\)</span>?</p>
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<input class="form-check-input" type="radio" name="radio_17874397323483167649" id="radio_17874397323483167649_1" value="1">
<span class="label-body px-1">
It is convergent
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_17874397323483167649_2">
<input class="form-check-input" type="radio" name="radio_17874397323483167649" id="radio_17874397323483167649_2" value="2">
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It is divergent
</span>
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</div>
<div class="form-check">
<label class="form-check-label" for="radio_17874397323483167649_3">
<input class="form-check-input" type="radio" name="radio_17874397323483167649" id="radio_17874397323483167649_3" value="3">
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Can't say
</span>
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</div>
</div>
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<hr>
<p>Let <span class="math inline">\(f(x) = 1 + 4x^2\)</span>. What can you say about <span class="math inline">\(\int_1^\infty f(x) dx\)</span>, as <span class="math inline">\(f(x) \leq 1/x^2\)</span> on <span class="math inline">\([1, \infty)\)</span>?</p>
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<input class="form-check-input" type="radio" name="radio_11890380834893891107" id="radio_11890380834893891107_1" value="1">
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It is convergent
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_11890380834893891107_2">
<input class="form-check-input" type="radio" name="radio_11890380834893891107" id="radio_11890380834893891107_2" value="2">
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It is divergent
</span>
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<div class="form-check">
<label class="form-check-label" for="radio_11890380834893891107_3">
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Can't say
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<div id="11890380834893891107_message" style="padding-bottom: 15px"></div>
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<hr>
<p>Let <span class="math inline">\(f(x) = \lvert \sin(x)^{10}\rvert/e^x\)</span>. What can you say about <span class="math inline">\(\int_1^\infty f(x) dx\)</span>, as <span class="math inline">\(f(x) \leq e^{-x}\)</span> on <span class="math inline">\([1, \infty)\)</span>?</p>
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<input class="form-check-input" type="radio" name="radio_15544107282915800719" id="radio_15544107282915800719_1" value="1">
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It is convergent
</span>
</label>
</div>
<div class="form-check">
<label class="form-check-label" for="radio_15544107282915800719_2">
<input class="form-check-input" type="radio" name="radio_15544107282915800719" id="radio_15544107282915800719_2" value="2">
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It is divergent
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Can't say
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<section id="question-9" class="level6">
<h6 class="anchored" data-anchor-id="question-9">Question</h6>
<p>The difference between “blowing up” at <span class="math inline">\(0\)</span> versus being integrable at <span class="math inline">\(\infty\)</span> can be seen to be related through the <span class="math inline">\(u\)</span>-substitution <span class="math inline">\(u=1/x\)</span>. With this <span class="math inline">\(u\)</span>-substitution, what becomes of <span class="math inline">\(\int_0^1 x^{-2/3} dx\)</span>?</p>
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<div class="cell-output cell-output-display" data-execution_count="23">
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<input class="form-check-input" type="radio" name="radio_14639808955046759817" id="radio_14639808955046759817_1" value="1">
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\(\int_1^\infty u^{2/3}/u^2 \cdot du\)
</span>
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<label class="form-check-label" for="radio_14639808955046759817_2">
<input class="form-check-input" type="radio" name="radio_14639808955046759817" id="radio_14639808955046759817_2" value="2">
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\(\int_0^\infty 1/u \cdot du\)
</span>
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</div>
<div class="form-check">
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<input class="form-check-input" type="radio" name="radio_14639808955046759817" id="radio_14639808955046759817_3" value="3">
<span class="label-body px-1">
\(\int_0^1 u^{2/3} \cdot du\)
</span>
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</div>
</div>
</div>
<div id="14639808955046759817_message" style="padding-bottom: 15px"></div>
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</section>
<section id="question-10" class="level6">
<h6 class="anchored" data-anchor-id="question-10">Question</h6>
<p>The antiderivative of <span class="math inline">\(f(x) = 1/\pi \cdot 1/\sqrt{x(1-x)}\)</span> is <span class="math inline">\(F(x)=(2/\pi)\cdot \sin^{-1}(\sqrt{x})\)</span>.</p>
<p>Find <span class="math inline">\(\int_0^1 f(x) dx\)</span>.</p>
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