# Problem of the Week #16 - July 16th, 2012

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#### Chris L T521

##### Well-known member
Staff member
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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Remember to read the POTW submission guidelines to find out how to submit your answers!

#### Chris L T521

##### Well-known member
Staff member
This week's question was correctly answered by Sudharaka. You can find his solution below.

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, $$\displaystyle\int_{-a}^{a}\frac{\sin x}{(x^2+1)^2}\,dx = 0$$.

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

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