How to prove that $\int_{a}^{+\infty}\int_{0}^{+\infty}e^{-xt}\sin t\,dx\,dt = \int_{0}^{+\infty}\frac{\cos a+x\sin a}{1+x^2}e^{-ax}\,dx$

Question

I want to prove $\int_{a}^{+\infty}\int_{0}^{+\infty}e^{-xt}\sin t\,dx\,dt = \int_{0}^{+\infty}\frac{\cos a+x\sin a}{1+x^2}e^{-ax}\,dx$.

Should the proof be done using some kinds of Laplace transform relations?

Answer 1

Change the order of integration, and do $$ \int_a^{\infty} e^{-xt} \sin{t} \, dt, $$ which, as you can see, is close to, but not quite, the Laplace transform of $\sin{t}$. What you need to do is one of: integrating by parts twice, or switching the sine into complex form and doing the integral directly. It should then fall out without much fuss.

Answer 2

Here is another way to show equality. Let

$$G(a)=\int_a^{\infty}\int_0^{\infty} e^{-xt}\sin t \,dx\,dt$$

and

$$H(a) = \int_0^{\infty} e^{-ax}\frac{\cos a+x\sin a}{1+x^2} \,dx.$$

Taking the derivatives of $G$ and $H$ reveals that

$$\begin{align} G'(a)&=-\int_0^{\infty} e^{-ax}\sin a \,dx\\\\ &=-\sin a\int_0^{\infty} e^{-ax}\,dx \end{align}$$

and

$$\begin{align} H'(a) &= \int_0^{\infty} e^{-ax}\frac{-x(\cos a+x\sin a)+(-\sin a+x\cos a)}{1+x^2} \,dx\\\\ &=\int_0^{\infty} e^{-ax}\frac{-(1+x^2)\sin a}{1+x^2} \,dx\\\\ &=-\sin a \int_0^{\infty} e^{-ax}\,dx. \end{align}$$

Thus, $G'(a) =H'(a)$, for all $a$ which implies that $G(a)=H(a)+C$, where C is a constant of integration.

But, for $a=0$, we have

$$\begin{align} G(0)&=\int_0^{\infty}\int_0^{\infty} e^{-xt}\sin t \,dx\,dt=\int_0^{\infty} \frac{\sin t}{t}\,dt\\\\ &=\pi/2 \end{align}$$

and

$$\begin{align} H(0) &= \int_0^{\infty} \frac{1}{1+x^2} \,dx\\\\ &=\pi/2 \end{align}$$

Thus, $G(a)=H(a)$ as was to be shown!

Answer 3

Changing the order of integration gives the integral

$$ \int_a^{\infty} e^{-xt} \sin t\,dt $$

which is equivalent to $$ \int_0^\infty e^{-xt} \sin t\,\operatorname{u}(t-a)\,dt = \mathcal{L}\left\{\sin t \operatorname{u}(t-a) \right\}(x) $$

where $\operatorname{u}(t)$ is the Heaviside step function $$\operatorname{u}(t-a) = \left\{ \begin{matrix} 0, & t < a \\ 1, & t > a \end{matrix} \right.$$

Using the shift property

$$ \begin{align} \mathcal{L}\left\{\sin t \operatorname{u}(t-a) \right\}(x) &= e^{-ax}\mathcal{L}\left\{ \sin(t+a)\right\}(x) \\[6pt] &= e^{-ax}\mathcal{L}\left\{\cos a \sin t + \sin a\cos t \right\}(x) \\[6pt] &= e^{-ax}\left(\cos a\cdot \frac{1}{1+x^2} + \sin a\cdot\frac{x}{1+x^2}\right) \\[6pt] &= \frac{\cos a + x\sin a}{1+x^2}\,e^{-ax} \end{align} $$