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Improper Integral

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Tolaso J Kos
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Improper Integral

#1

Post by Tolaso J Kos » Fri Jul 08, 2016 1:06 pm

Prove that: \( \displaystyle \int_{-\infty }^{\infty }\frac{\cos x}{x^2+a^2}\, dx=\frac{\pi e^{-a}}{a} \) where \(a>0 \)
Imagination is much more important than knowledge.
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Re: Improper Integral

#2

Post by admin » Fri Jul 08, 2016 1:07 pm

Replied by ex-member aziiri:

Consider \(g(z)=\frac{e^{i z}}{z^2+a^2}\), and let \(C_r\) be a the half circle \(\{re^{i t} | 0\leq t\leq \pi\}\) with \(r>a\), then : \[\int_{C_r \cup [-r,r]} g(z) \ \mathrm{d}z = 2\pi i \text{Res}(g,ia)= \frac{2\pi i e^{-a}}{2ia}= \frac{\pi e^{-a} }{a}\] And by the estimation lemma we have: \[\left|\int_{C_r} g(z) \ \mathrm{d}z \right|\leq \int_{C_r} \left|\frac{e^{i z}}{z^2+a^2}\right| \ \mathrm{d}z \leq \int_{C_r} \frac{\mathrm{d}z}{r^2-a^2} = \frac{\pi r}{r^2-a^2}\] Now, take \(r\to \infty\) to get : \[\int\limits_{-\infty}^{+\infty}g(z) \ \mathrm{d}z = \frac{\pi e^{-a} }{a}.\] And therefore : \[\int\limits_{-\infty}^{+\infty}\frac{\cos x}{x^2+a^2} \ \mathrm{d}x= \text{Re}\left(\int\limits_{-\infty}^{+\infty}g(z) \ \mathrm{d}z\right) = \frac{\pi e^{-a}}{a}.\]
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Grigorios Kostakos
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Re: Improper Integral

#3

Post by Grigorios Kostakos » Fri Jul 08, 2016 1:08 pm

A 2nd solution with real analysis methods:

We use that \begin{align*}
\displaystyle \int_{0}^{+\infty }{2t\,{\mathrm{e}}^{-(x^2+a^2)t^2}\, dt}&=\frac{1}{x^2+a^2}\qquad&(1)\\
\displaystyle \int_{0}^{+\infty }{{\mathrm{e}}^{-t^2x^2}\cos{x}\, dx}&=\frac{\sqrt{\pi}}{2t}\,{\mathrm{e}}^{-\frac{1}{4t^2}}\qquad&(2)\\
\int_{0}^{+\infty}{{\mathrm{e}}^{-a^2t^2-\frac{1}{4t^2}}\, dt}&=\frac{\sqrt{\pi}}{2a}\,{\mathrm{e}}^{-a}\qquad&(3)
\end{align*}So
\begin{align*}
\displaystyle \int_{-\infty }^{+\infty }{\frac{\cos x}{x^2+a^2}\, dx}&=\int_{-\infty }^{0}{\frac{\cos x}{x^2+a^2}\, dx}+\int_{0}^{+\infty }{\frac{\cos x}{x^2+a^2}\, dx}\\
&\mathop{=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{u\,=\,-x}\\
{-du\,=\,dx}\\
\end{subarray}}\,\int_{0}^{+\infty}{\frac{\cos u}{u^2+a^2}\, du}+\int_{0}^{+\infty }{\frac{\cos x}{x^2+a^2}\, dx}\\
&=2\int_{0}^{+\infty }{\frac{\cos x}{x^2+a^2}\, dx}\\
&\stackrel{(1)}{=\!=}2\int_{0}^{+\infty }{\biggl({\int_{0}^{\infty }{2t\,{\mathrm{e}}^{-(x^2+a^2)t^2}\, dt}}\bigg)\cos x\, dx}\\
&=2\int_{0}^{+\infty }{2t\,{\mathrm{e}}^{-a^2t^2}\biggl({\int_{0}^{+\infty }{{\mathrm{e}}^{-x^2t^2}\cos x\,dx}}\bigg)\, dt}\\
&\stackrel{(2)}{=\!=}2\int_{0}^{+\infty }{2t\,{\mathrm{e}}^{-a^2t^2}\frac{\sqrt{\pi}}{2t}{\mathrm{e}}^{-\frac{1}{4t^2}}\, dt}\\
&=2\sqrt{\pi}\int_{0}^{+\infty}{{\mathrm{e}}^{-a^2t^2-\frac{1}{4t^2}}\, dt}\\
&\stackrel{(3)}{=\!=}2\sqrt{\pi}\,\frac{\sqrt{\pi}}{2a}\,{\mathrm{e}}^{-a}\\
&=\frac{\pi \,{\mathrm{e}}^{-a}}{a}\,.
\end{align*}

Proofs of (1), (2) \(\&\) (3):

(1) Trivial.

(2) We will prove the more general \ Differentiating with respect to \(k\), by Leibnitz rule, we have \begin{align*}
\frac{dI}{dk}&=-\displaystyle \int_{0}^{+\infty }{x\,{\mathrm{e}}^{-t^2x^2}\sin({kx})\, dx}\\
&= \frac{1}{2t^2}\int_{0}^{+\infty }{\bigl({{\mathrm{e}}^{-t^2x^2}}\bigr)'\sin({kx})\, dx}\\
&=\biggl[{\frac{1}{2t^2}\,{\mathrm{e}}^{-t^2x^2}\sin({kx})}\biggr]_{0}^{+\infty }-\frac{k}{2t^2}\int_{0}^{+\infty }{x\,{\mathrm{e}}^{-t^2x^2}\cos({kx})\, dx}\\
&=0-\frac{k}{2t^2}\,I\qquad\Rightarrow\\
\frac{dI}{dk}&=-\frac{k}{2t^2}\,I\qquad\Rightarrow\\
\frac{dI}{I}&=-\frac{k}{2t^2}\,dk\qquad\Rightarrow\\
\log{I}&=-\frac{k^2}{4t^2}+c\quad\Rightarrow\\
I&=c_1{\mathrm{e}}^{-\frac{k^2}{4t^2}}\,.
\end{align*} For \(k=0\) we have that \[\displaystyle c_1{\mathrm{e}}^{-\frac{0^2}{4t^2}}=\int_{0}^{+\infty }{{\mathrm{e}}^{-t^2x^2}\cos(0\cdot x)\, dx}= \int_{0}^{+\infty }{{\mathrm{e}}^{-t^2x^2}\, dx}=\frac{\sqrt{\pi}}{2t}\,,\] and we get \(c_1=\frac{\sqrt{\pi}}{2t}\,.\) So we have the desired result.

(3) Left as an exercise.
Grigorios Kostakos
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