Solving exponential integral with two variables (x and r)

Question

Prove that $\int_0^\infty \frac{x^r}{x\sqrt{2\pi}}e^{\frac{-ln(x)^2}{2}}dx=e^\frac{r^2}{2}$

I tried integrating by parts, with $u=e^{\frac{-ln(x)^2}{2}}$ and $v' =$ the remaining. Thanks.

Answer 1

\Use the replacement $\log x = t$, then $x = \mathrm{e}^t$ and $dt = dx/x$, so your integral becomes $$ \int_{-\infty}^\infty e^{r t} \frac{e^{-t^2/2}}{\sqrt{2\pi}} dt $$

In which you should recognize a standard normal distribution. Apply rules for Gaussian integration and you're done!

EDIT: Here is how to solve the final integral. A neat trick in Gaussian integration. $$\int_{-\infty}^{\infty} \frac{e^{-t^2/2 + rt}}{\sqrt{2\pi}}dt $$ Complete the square at the exponent as $-t^2/2 + rt +r^2/2 - r^2/2 = -(t - r)^2/2 + r^2/2$, we get $$\int_{-\infty}^{\infty} \frac{e^{-(t-r)^2/2 + r^2/2}}{\sqrt{2\pi}}dt =e^{r^2/2}\int_{-\infty}^{\infty} \frac{e^{-(t-r)^2/2}}{\sqrt{2\pi}}dt$$ Note that what is left is the total area under a Gaussian density with mean r and variance 1, so the integral is $1$. QED