This question is from my university paper:
$$\int_{0}^{\infty} \int_{0}^{x} x e^{\frac{-x^2}{y}}\; dy\;dx = \frac{1}{2}$$
I tried using DUIS method but integration is getting complicated
$f(x) = \int_{0}^{x} x e^{\frac{-x^2}{y}}\; dy$
$ \frac{d}{dx}f(x) = \int_{0}^{x} \frac{d}{dx} x e^{\frac{-x^2}{y}}\; dy $
$ \frac{d}{dx}f(x) = \int_{0}^{x} \frac{e^{-x^2}(1-2x^2)}{y}\; dy $
$ \hspace{75px}= e^{(-x^2)} (1 - 2 x^2) \{\log(x)-\log(0)\}$
$ \hspace{75px}= e^{(-x^2)} (1 - 2 x^2)\;\log(x)$
$\int \frac{d}{dx}f(x) = \int{e^{(-x^2)} (1 - 2 x^2)\;\log(x) \;dx} $
First we must integrate with respect to $y$ and that's where I'm getting stuck.
Is there any other method to solve this? What's going wrong?
Why don't you do what is written in your question's title?:
$$\int_0^\infty\int_0^x xe^{-x^2/y}\,dy\,dx=\int_0^\infty\int_y^\infty xe^{-x^2/y}\,dx\,dy=\int_0^\infty\left.\left(-\frac y2\right)e^{-x^2/y}\right|_y^\infty dy=$$
$$=-\frac12\int_0^\infty y\left(0-e^{-y}\right)\,dy=\frac12\int_0^\infty ye^{-y}dy=\left.\frac12\left[-ye^{-y}\right]\right|_0^\infty+\frac12\int_0^\infty e^{-y}dy=$$
$$\left.\frac12\left(-e^{-y}\right)\right|_0^\infty=\frac12$$