a question about integral with parameter variables?

Question

I have a problem proving $$\int_{0}^\infty dx {\left(\int_{0}^\infty e^{-x^2t}\sin t\, dt\right)}=\int_{0}^\infty dt\left( \int_{0}^\infty e^{-x^2t}\sin t\, dx\right)$$.

I have been struggling for it for a long time, I don't know how to prove it. Or, maybe it is wrong? Can somebody tell me how to prove it? Why they can change the order of integration?

Answer 1

Here is a more elementary way of doing this.

\begin{align} \int_{0}^\infty \int_{0}^\infty e^{-x^2t}\sin t \;dt dx&= \int_{0}^\infty \mathcal{L}\{ \sin(t) \} \;dx\\ &=\int_{0}^\infty \frac{1}{(x^2)^2 + 1} dx \quad {\text{You are going to have to use partial fractions and a few substitutions. }}\\ &= \frac{\pi}{2\sqrt{2}} \end{align}

Answer 2

Since you have the integrals specifically, why not just evaluate them? On the LHS

$$\int_0^{\infty} dt \, e^{-x^2 t} \sin{t} = \operatorname{Im} \left [\int_0^{\infty} dt \, e^{-(x^2-i) t} \right ] = \operatorname{Im} \left [\frac1{x^2-i} \right ] = \frac1{x^4+1}$$

$$\int_0^{\infty} \frac{dx}{x^4+1} = \frac{i 2 \pi}{(1-i) (4 e^{i 3 \pi/4})} = \frac{\pi}{2 \sqrt{2}}$$

On the RHS:

$$\int_0^{\infty} dx \, e^{-x^2 t} = \frac12 \sqrt{\frac{\pi}{t}}$$

$$\frac{\sqrt{\pi}}{2} \int_0^{\infty} dt \, \frac{\sin{t}}{\sqrt{t}} = \frac{\sqrt{\pi}}{2}\int_{-\infty}^{\infty} du \, \sin{u^2} = \frac{\pi}{2 \sqrt{2}}$$

Note:

$$\int_{-\infty}^{\infty} du \, \sin{u^2} = 2 \operatorname{Im} \left [\int_{0}^{\infty} du \, \, e^{i u^2} \right ] = 2 \operatorname{Im} \left [e^{i \pi/4}\int_{0}^{\infty} dv \, \, e^{-v^2} \right ] = \frac{\sqrt{\pi}}{\sqrt{2}}$$

So, there you go, you can reverse the order of integration.