# Thread: [SOLVED] Finding the integration result

1. ## [SOLVED] Finding the integration result

Actually I was trying to find the expected value, nevertheless, hope someone can give me some idea of deriving the result:

given: $\displaystyle f(x|\theta)=\frac{x}{\theta ^2}e^{(-x^2)/(2\theta ^2)}$ $\displaystyle x\geq 0$

I want to solve for: $\displaystyle \int_{0}^{\infty}\frac{x^2}{\theta ^2}e^{(-x^2)/(2\theta ^2)}.dx$ .

2. Originally Posted by noob mathematician
Actually I was trying to find the expected value, nevertheless, hope someone can give me some idea of deriving the result:

given: $\displaystyle f(x|\theta)=\frac{x}{\theta ^2}e^\frac{-x^2}{(2\theta ^2)}$ $\displaystyle x\geq 0$

I want to solve for: $\displaystyle \int_{0}^{\infty}\frac{x^2}{\theta ^2}e^\frac{-x^2}{(2\theta ^2)}$ .
Make the substitution $\displaystyle u=x^2/(2\theta^2)$.

Then $\displaystyle \,du=x/ \theta^2\,dx$.

So $\displaystyle \int_0^{\infty}\frac{x^2}{\theta^2}e^{-x^2/(2\theta^2)}\,dx=\sqrt{2}\theta\int_0^{\infty}u^{1/2}e^{-u}\,du=\sqrt{2}\theta\Gamma\!\left(\tfrac{3}{2}\ri ght)=\tfrac{1}{2}\sqrt{2}\theta\Gamma\!\left(\tfra c{1}{2}\right)=\tfrac{1}{2}\sqrt{2\pi}\theta$.

Does this make sense?

3. Originally Posted by Chris L T521
Make the substitution $\displaystyle u=x^2/(2\theta^2)$.

Then $\displaystyle \,du=x/ \theta^2\,dx$.

So $\displaystyle \int_0^{\infty}\frac{x^2}{\theta^2}e^{-x^2/(2\theta^2)}\,dx=\sqrt{2}\theta\int_0^{\infty}u^{1/2}e^{-u}\,du=\sqrt{2}\theta\Gamma\!\left(\tfrac{3}{2}\ri ght)=\tfrac{1}{2}\sqrt{2}\theta\Gamma\!\left(\tfra c{1}{2}\right)=\tfrac{1}{2}\sqrt{2\pi}\theta$.

Does this make sense?
Thanks make full sense!