show $\int_0^{\infty} \frac{x\sin ax}{x^2+t^2}dx = \frac{\pi}{2}e^{-at}$

Question

I'm asked to show $\int_0^{\infty} \frac{x\sin ax}{x^2+t^2}dx = \frac{\pi}{2}e^{-at}$ when $t,a > 0$. I tried using integration by parts integrating $\frac{x}{x^2+t^2}$. But it seems that $\frac{x\sin ax}{x^2+t^2}$ has no antiderivative. What should I try?

Answer 1

Using contour integration or Fourier inversion, one can show that if $a>0$ then $$ \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{e^{iax}}{1+x^2}\;dx=e^{-a}$$ which implies that $$ \frac{2}{\pi}\int_{0}^{\infty}\frac{\cos(ax)}{1+x^2}\;dx=e^{-a}$$ if $a>0$. Differentiating with respect to $a$ then yields $$\int_{0}^{\infty}\frac{x\sin(ax)}{1+x^2}\;dx=\frac{\pi}{2}e^{-a}$$ and finally replacing $a$ with $at$ and making a substitution in the integral yields the desired result.

Answer 2

By residue theorem, and due to the parity of your integrand, we can write it as

$$\frac{1}{2}\int_{-\infty}^{+\infty} \frac{z\Im \left(e^{iaz}\right)}{(z + it)(z-it)}\ dz = \frac{1}{2}\Im\int_{|z|=1} \frac{z\Im \left(e^{iaz}\right)}{(z + it)(z-it)}\ dz$$

$x = +it$ lies in the upper half plane hence by residues formula:

$$\Im \frac{1}{2}2\pi i \lim_{z\to it} (z-it)\frac{z e^{iaz}}{(z+it)(z-it)} = \Im \frac{\pi}{2}i e^{-at}$$

Whence the result

$$\color{red}{\frac{\pi}{2}e^{-at}}$$

Answer 3

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I thought it might be instructive to present a way forward that relies on real analysis only and forgoes complex analysis. To that end, we now proceed.

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Let $I(a,t)$ be given by

$$\begin{align} I(a,t)&=\int_0^\infty \frac{x\sin(ax)}{x^2+t^2}\,dx\\\\ &\overbrace{=}^{x\mapsto x|t|}\int_0^\infty \frac{x\sin(a|t|x)}{x^2+1}\,dx\\\\ &=\int_0^\infty \frac{(x^2+1-1)\sin(a|t|x)}{x(x^2+1)}\,dx\\\\ &=\int_0^\infty \frac{\sin(a|t|x)}{x}\,dx-\int_0^\infty \frac{\sin(a|t|x)}{x(x^2+1)}\,dx\\\\ &=\frac\pi2\text{sgn}(a|t|)-\int_0^\infty \frac{\sin(a|t|x)}{x(x^2+1)}\,dx\tag1 \end{align}$$

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For $|at|>\delta>0$, the integrals $\int_0^\infty \frac{\cos(a|t|x)}{x^2+1}\,dx$ and $\int_0^\infty \frac{x\sin(a|t|x)}{x^2+1}\,dx$ uniformly converge and we may differentiate twice (with respect to $a|t|$) under the integral on the right-hand side of $(1)$ to find for $|at|\ge \delta>0$

$$\frac{d^2I(a,t)}{d(a|t|)^2}=I(a,t)\tag2$$

The general solution to $(2)$ is $I(a,t)=Ae^{-a|t|}+Be^{a|t|}$. To find the integration constants $A$ and $B$ we invoke the conditions (i) $\lim_{at\to 0^+}I(a,t)=\frac{\pi}{2}$ and $(ii)$ $\left.\left(\frac{dI(a,t)}{d(a|t|)}\right)\right|_{a|t|=0}=-\frac\pi2$. Proceeding, we find that $A=\frac\pi2$ and $B=0$.

Putting it all together and exploiting the oddness of the integrand around $a$, we find that

$$I(a,t)=\text{sgn}(a)\frac\pi2 e^{-|at|}$$

And we are done!

Answer 4

Let’s first evaluate the integral using contour integration along anti-clockwise direction of the path $$\gamma=\gamma_{1} \cup \gamma_{2} \textrm{ where } \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} (0<t<\pi) $$ As $R \rightarrow +\infty$, we have $$ \begin{aligned}\int_0^{\infty} \frac{\cos (a x)}{x^2+t^2} d x&=\frac{1}{2}\left(\int_{-\infty}^{\infty} \frac{e^{a x i}}{x^2+t^2} d z\right) \\ & =\frac{1}{2} \Re \int_\gamma \frac{e^{a zi}}{z^2+t^2} d x \\ & =\frac{1}{2} \Re\left[2 \pi i \operatorname{lim}_{z \rightarrow ti}(z-t i) \frac{e^{a z i}}{z^2+t^2} d z\right] \\ & =\pi \Re\left(i \cdot \frac{e^{a(t i) i}}{2 t i}\right)\\&=\frac{\pi}{2 t} e^{-a t} \end{aligned} $$

Differentiating both sides w.r.t. $a$ yields our integral $$ \boxed{\int_0^{\infty} \frac{x \sin (a x)}{x^2+t^2} d x=\frac{\pi}{2} e^{-a t}} $$