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      <text>13.2.3.1 Infinite limits</text>
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      <text><![CDATA[<h3>Some integrals with infinite limits</h3>
<p>On occasions, and notably when dealing with Laplace and Fourier transforms, you will come across integrals in which one of the limits is infinite. We avoid a rigorous treatment of such cases here and instead give some commonly occurring examples.</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Find the definite integral of \(e^{-x}\) from \(0\) to \(\infty\); that is, find \({\displaystyle\int_0^\infty e^{-x} \, \mathrm{d}x}\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>The integral is found in the normal way:</p>

\({\displaystyle\int_0^\infty e^{-x} \, \mathrm{d}x}\) = \(\bigg[- e^{-x}\bigg]_0^\infty\)

<p>There is no difficulty in evaluating the square bracket at the lower limit. We obtain simply \(-e^{-0}=-1\). At the upper limit we must examine the behaviour of \(-e^{-x}\) as \(x\) gets infinitely large. This is where it is important that you are familiar
    with the properties of the exponential function. If you refer to the graph (see the figure below) you will see that as \(x\) tends to infinity \(e^{-x}\) tends to zero.</p>
<figure class="HELM_figure"><img src="@@PLUGINFILE@@/13.2%20Graph%20of%20e%5E%28-x%29.svg" alt="The graph of e to the negative x decreases towards zero for increasing x." width="307" height="148" class="img-responsive atto_image_button_text-bottom"></figure>

<p>Consequently the contribution to the integral from the upper limit is zero. So</p>
<p>\({\displaystyle\int_0^\infty e^{-x} \, \mathrm{d}x}\) = \(\left[\phantom{\dfrac{1}{1}} - e^{-x}\right]_0^\infty\) = \((-e^{-\infty})-(e^{-0})\) = \((0) - (e^{-0})\) = \(1\)</p>

<p>Thus the value of \({\displaystyle\int_0^\infty e^{-x} \, \mathrm{d}x}\) is 1.</p>

<p>Another way of achieving this result is as follows: We change the infinite limit to a finite limit, \(b\), say and then examine the behaviour of the integral as \(b\) tends to infinity, written as \({\displaystyle\int_0^\infty e^{-x} \, \mathrm{d}x} =
    {\displaystyle\lim_{b\rightarrow\infty}}{\displaystyle\int_0^b e^{-x} \, \mathrm{d}x}\)</p>
<p>Now,</p> \({\displaystyle\int_0^b e^{-x} \, \mathrm{d}x} = \left[\phantom{\dfrac{1}{1}}e^{-x}\right]_0^b = \left( -e^{-b}\right) - \left(-e^{-0}\right) = -e^{-b} + 1\)
<p></p>

<p>Then as \(b\) tends to infinity \(-e^{-b}\) tends to zero, and the resulting integral has the value 1, as before.</p>

<p>Many integrals having infinite limits cannot be evaluated in a simple way like this, and many cannot be evaluated at all. Fortunately, most of the integrals you will meet will exhibit the sort of behaviour seen in the last example.</p>]]></text>
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