Integral Trouble!!!

[smithsonian7]

Show that the \(\displaystyle \Large\int_{0}^{\infty}\frac{x}{z^{2}}e^{\frac{-x^{2}}{2z^{2}}}dx = -e^{\frac{-x^{2}}{2z^{2}}}=1\).

I am using a solution manual to complete a chapter from a probability and statistics text book (Probability and Statistics for Engineering and the Sciences, Devore, Sixth Edition, Ch. 4, question 4) in preparation for the actuary exam on probability. The solution manual gave me the above end result but failed to demonstrate how they got it. I don't like to move past things I don't understand so if any help in solving this integral step by step would be greatly appreciated. I thought that it originally used integration by parts since its the integral of f(x)*g(x) but I couldn't get it to look like the solution they found.

Thanks

 

[galactus]

I erroneously stated in my last post that this was not integrable by elementary means. I was incorrect.

That's what I get for just glancing. I was thinking about the error function, which is used in stats and is not integrable by the usual means.

But, continuing, this is a rather easy integration:

Let \(\displaystyle u=\frac{-x^{2}}{2z^{2}}, \;\ -du=\frac{x}{z^{2}}dx\)

Make the subs and it whittles down to a simple integration.

\(\displaystyle -\int e^{u}du=-e^{u}\)

Resub:

\(\displaystyle -e^{\frac{-x^{2}}{2z^{2}}}\)

You may have to treat this as an improper integral because of the infinity. Just plugging it in after integrating may result in an undefined result. But allowing the limit to head toward infinity, we get 0. Because \(\displaystyle -e^{\frac{-x^{2}}{2z^{2}}}=\frac{-1}{e^{\frac{x^{2}}{2z^{2}}}}\). Of course, the 1 evaluates to -1.

Then, we get 0-(-1)=1