Contour integral calculation 2

Question

Show that \begin{align*} \frac{1}{a}\int_0^{\infty} \frac{\sin x}{x+a} \; dx= \int_0^{\infty} \frac{e^{-x}}{x^2+a^2}\;dx \quad (a>0) \end{align*}

My Attempt: I showed that

\begin{align*} \frac{1}{a}\int_0^{\infty} \frac{\sin x}{x+a} \; dx &= \frac{1}{2ai} \int_0^{\infty} \frac{e^{ix}-e^{-ix}}{x+a} \;dx \\ &= \frac{1}{2ai}\int_0^{\infty} \frac{e^{-x}-e^x}{x-ia} \; dx \end{align*} by using contur integral. Also, \begin{align*} \int_0^{\infty} \frac{e^{-x}}{x^2+a^2}\;dx =\frac{1}{2ai}\int_0^{\infty} \frac{e^{-x}}{x-ai}-\frac{e^{-x}}{x+ai} \; dx \end{align*} Thus, It suffices to show that \begin{align*} \int_0^{\infty} \frac{e^x}{x-ia} \; dx = \int_0^\infty \frac{e^{-x}}{x+ai} \; dx \end{align*}

But, I stuck how to prove that....

Any help is appreciated.

Answer 1

You can only perform contour integration around infinite region if the part that circles around infinity vanishes. Otherwise you get divergent integrals (your integrals that include $e^{+x}$ cannot possibly converge). Your integral:

$$\frac1{2ai}\int_0^\infty \frac{e^{ix}-e^{-ix}}{x+a}dx$$ consists of two integrands. The first term $e^{ix}$ will converge to zero on the upper half-plane. For it, you can integrate $0+0i\to \infty+0i \to 0+\infty i \to 0+0i$. You also know you did not encircle any poles, so you can write: $$\frac1{2ai}\int_0^\infty \frac{e^{ix}}{x+a}dx+\frac1{2ai}\int_{\infty i}^0 \frac{e^{ix}}{x+a}dx=0$$ substitute $x=0+it$, reverse integration boundaries and carry to the right side of the equals sign: $$\frac1{2ai}\int_0^\infty \frac{e^{ix}}{x+a}dx=\frac1{2a}\int_0^\infty \frac{e^{-t}}{it+a} dt$$

The second integral converges in the lower half-plane, so you must integrate $0+0i\to \infty + 0i \to 0-\infty i \to 0+0i$ to get something that includes your integral and has a zero contribution on the quarter-circle infinite arc from real to imaginary infinity. $$\frac1{2ai}\int_0^\infty \frac{e^{-ix}}{x+a}dx+\frac1{2ai}\int_{-\infty i}^0 \frac{e^{-ix}}{x+a}dx=0$$ Now, use $x=0-it$ and you get: $$\frac1{2ai}\int_0^\infty \frac{e^{-ix}}{x+a}dx=-\frac1{2a}\int_0^{\infty} \frac{e^{-t}}{-it+a}dt$$

Now (once you are sure they both individually converge), you can put them together (there is a minus between the terms): $$\frac1{2ai}\int_0^\infty \frac{e^{ix}-e^{-ix}}{x+a}dx=\frac{1}{2a}\int_0^\infty e^{-t}\left(\frac{1}{it+a}+\frac{1}{-it+a}\right)dt$$ You can do the final recombination of the fractions yourself.

Answer 2

The Laplace transform is a self-adjoint operator, i.e. $$ \int_{0}^{+\infty} f(x)\cdot(\mathcal{L} g)(x)\,dx = \int_{0}^{+\infty} g(x)\cdot(\mathcal{L} f)(x)\,dx$$ by Fubini's theorem. For any $a>0$ we have $$ \mathcal{L}(\sin x)(s) = \frac{1}{s^2+1},\qquad \mathcal{L}^{-1}\left(\frac{1}{x+a}\right)=e^{-ax} $$ hence $$ \int_{0}^{+\infty}\frac{\sin x}{a(x+a)}\,dx = \int_{0}^{+\infty}\frac{e^{-ax}}{a(x^2+1)}\,dx\stackrel{x\mapsto t/a}{=}\int_{0}^{+\infty}\frac{e^{-t}}{t^2+a^2}\,dt$$ QED.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{{1 \over a}\int_{0}^{\infty}{\sin\pars{x} \over x + a} \,\dd x = \int_{0}^{\infty}{\expo{-x} \over x^{2} + a^{2}}\,\dd x:\ {\LARGE ?} \qquad \pars{~a > 0~}}$.

With $\ds{R > a > 0}$: \begin{align} &\bbox[10px,#ffd]{\left. {1 \over a}\int_{0}^{R}{\sin\pars{x} \over x + a} \,\dd x\,\right\vert_{\ a\ >\ 0}} = {1 \over a}\,\Im\int_{0}^{R}{\expo{\ic x} \over x + a} \,\dd x \\[5mm] = &\ -\,{1 \over a}\,\Im\int_{0}^{\pi/2}{\exp\pars{\ic R\expo{\ic\theta}} \over R\expo{\ic\theta} + a} \,R\expo{\ic\theta}\ic\,\dd\theta - {1 \over a}\,\Im\int_{R}^{0}{\expo{\ic\pars{\ic y}} \over \ic y + a}\,\ic\,\dd y \\[5mm] = &\ -\,{R \over a}\,\Re\int_{0}^{\pi/2}{\exp\pars{\ic R\expo{\ic\theta}} \over R\expo{\ic\theta} + a} \,\expo{\ic\theta}\,\dd\theta + {1 \over a}\,\Re\int_{0}^{R}{a - \ic y \over y^{2} + a^{2}}\,\expo{-y}\dd y\label{1}\tag{1} \end{align}

Note that

\begin{align} & 0 < \verts{-\,{R \over a}\int_{0}^{\pi/2}{\exp\pars{\ic R\expo{\ic\theta}} \over R\expo{\ic\theta} + a} \,\expo{\ic\theta}\,\dd\theta} \\[5mm] < &\ {R \over a}\int_{0}^{\pi/2} {\exp\pars{-R\sin\pars{\theta}} \over R} \,\dd\theta \\[5mm] < & {1 \over a}\int_{0}^{\pi/2} \exp\pars{-2R\theta/\pi}\,\dd\theta = {1 \over a}\,{\expo{-R} - 1 \over 2R/\pi} \,\,\,\stackrel{\mrm{as}\ R\ \to\ \infty}{\to}\,\,\, 0 \label{2}\tag{2} \end{align}

\eqref{1} and \eqref{2}

$$ \implies \bbx{\left.{1 \over a}\int_{0}^{\infty}{\sin\pars{x} \over x + a} \,\dd x\,\right\vert_{\ a\ >\ 0} = \int_{0}^{\infty}{\expo{-x} \over x^{2} + a^{2}}\,\dd x} $$