a question about integral with parameter variables? 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?
 A: 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.
A: 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}
