Expectation of exponential of normal random variable

Question

I'm trying to work out the expectation of $E[e^{uX^{2}}]$ where $X \sim N(0, 1)$. Now, from my understanding: $$ E[e^{uX^{2}}] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{\frac{-x^{2}}{2}(1 - 2u)} = e^{(1 - 2u)}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{\frac{-x^{2}}{2}} = e^{1 - 2u} $$ However, my teacher found the result that $E[e^{uX^{2}}] = (1 - 2u)^{-\frac{1}{2}}$. However, I'm not sure how he obtained that result

Answer 1

Your step after the integral $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{\frac{-x^{2}}{2}(1 - 2u)}\,dx\tag1$$ is incorrect. To evaluate (1) you should make the change of variables $t:=x\sqrt{1-2u}$. Then $dt=dx\sqrt{1-2u}$ and (1) becomes $$ \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12t^2}\frac1{\sqrt{1-2u}}\,dt= \frac1{\sqrt{1-2u}}\underbrace{\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12t^2}\,dt}_1. $$