Is Contour Integration the Key to Solving This Integral?

[Chris L T521]

Thanks to those who participated in last week's POTW! Here's this week's problem.

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Problem: Using contour integration, show that

\[\int_{-\infty}^{\infty}\frac{\cos x}{(x^2+1)^2}\,dx = \frac{\pi}{e}.\]

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[Chris L T521]

This week's question was correctly answered by Sudharaka. You can find his solution below.

Spoiler

Let us consider the contour integral, \(\displaystyle\int_{C}\frac{e^{iz}}{(z^2+1)^2}\) where \(C\) is the semicircle that bounds the upper half of the disk of radius \(a>1\) centered at the origin.

\[\int_{C}\frac{e^{iz}}{(z^2+1)^2}\,dz=\int_{C} \frac{\frac{e^{iz}}{(z+i)^2}}{(z-i)^2}\,dz\]Since both \(e^{iz}\) and \(\displaystyle\frac{1}{(z+i)^2}\) are holomorphic on and inside the contour \(C\), \(\displaystyle\frac{e^{iz}}{(z+i)^2}\) is also holomorphic on and inside \(C\). Also \(i\) lies within \(C\). Therefore by Cauchy's Integral formula we get,\begin{eqnarray}\int_{C}\frac{e^{iz}}{(z^2+1)^2}\,dz&=&2\pi i\frac{d}{dz}\left[\frac{e^{iz}}{(z+i)^2}\right]_{z=i}\\&=&\frac{\pi}{e}\\\end{eqnarray}Note that, \(\displaystyle\int_{C}\frac{e^{iz}}{(z^2+1)^2}\,dz=\int_{-a}^{a}\frac{e^{iz}}{(z^2+1)^2}\,dz+\int_{arc}\frac{e^{iz}}{(z^2+1)^2}\,dz\) where \(arc\) is the arc of the semicircle \(C\).\[\therefore\int_{-a}^{a}\frac{e^{iz}}{(z^2+1)^2}\,dz=\frac{\pi}{e}-\int_{arc}\frac{e^{iz}}{(z^2+1)^2}\,dz~~~~~~~~~~~(1)\]On the semicircular arc, \(z=a(\cos\theta+i\sin\theta)\mbox{ where }0\leq\theta\leq\pi\mbox{ is the argument of }z.\)\begin{eqnarray}\therefore|e^{iz}|&=&|e^{ia(\cos\theta+i\sin\theta)}|\\&=&e^{-a\sin\theta}~~~~~~~~~~~~(2)\\\end{eqnarray}Since \(a>1\) by the reverse triangle inequality we get,\[|z^2+1|^2\geq||z|^2-1|^2=(a^2-1)^2\]\[\Rightarrow\frac{1}{|z^2+1|}\leq\frac{1}{(a^2-1)^2}~~~~~~~~~~~~~(3)\]By (2) and (3),\[\left|\frac{e^{iz}}{(z^2+1)^2}\right|\leq\frac{e^{-a\sin\theta}}{(a^2-1)^2}~~~~~~~~~~~~(4)\]Applying the Estimation lemma to, \(\displaystyle\int_{arc}\frac{e^{iz}}{(z^2+1)^2} \,dz\) and using (4) we get,\[\left|\int_{arc}\frac{e^{iz}}{(z^2+1)^2}\,dz\right|\leq\frac{\pi ae^{-a\sin\theta}}{(a^2-1)^2}\mbox{ where }a>1\mbox{ and }0\leq\theta\leq\pi\] Therefore by the Squeeze theorem,\[\lim_{a\rightarrow\infty}\int_{arc}\frac{e^{iz}}{(z^2+1)^2}\,dz=0~~~~~~~~~~~(5)\]By (1) and (5),\[\lim_{a\rightarrow\infty}\int_{-a}^{a}\frac{e^{iz}}{(z^2+1)^2}\,dz=\frac{\pi}{e}\]Using the Euler's formula we get,
\[\lim_{a\rightarrow\infty}\left[\int_{-a}^{a}\frac{\cos x}{(x^2+1)^2}\,dx+i\int_{-a}^{a}\frac{\sin x}{(x^2+1)^2}\,dx\right] = \frac{\pi}{e}\]Since, \(\displaystyle\frac{\sin x}{(x^2+1)^2}\) is an odd function, \(\displaystyle\int_{-a}^{a}\frac{\sin x}{(x^2+1)^2}\,dx = 0\).\[\therefore\int_{-\infty}^{\infty}\frac{\cos x}{(x^2+1)^2}\,dx = \frac{\pi}{e}\]