I am trying to work out the following integral:
$\displaystyle
\int_{0}^{\infty}xe^{ax -bx^2} dx
$
I have tried making several substitutions e.g. $\displaystyle u = ax -bx^2$ but have got stuck each time. Any help greatly appreciated
Cheers
I am trying to work out the following integral:
$\displaystyle
\int_{0}^{\infty}xe^{ax -bx^2} dx
$
I have tried making several substitutions e.g. $\displaystyle u = ax -bx^2$ but have got stuck each time. Any help greatly appreciated
Cheers
You'll want to complete the square in the exponent: use $\displaystyle u=x-\frac{a}{2b}$. Then $\displaystyle x=u+\frac{a}{2b}$, and the integral becomes:
$\displaystyle \int_{0}^{\infty}xe^{ax-bx^2}\,dx=\int_{-\frac{a}{2b}}^{\infty}\left(u+\frac{a}{2b}\right)e ^{\frac{a^2}{4b}-bu^2}\,du$
$\displaystyle \,\,=e^{\frac{a^2}{4b}}\int_{-\frac{a}{2b}}^{\infty}\left(u+\frac{a}{2b}\right)e ^{-bu^2}\,du$
Does that help?
--Kevin C.
What are the chances .....?? Virtually the exact same question was asked here a while back. I love it when you can recycle ....
I wonder if your question is NOT what you've posted, but some other question like ....... finding the moments of a distribution?
Hmmmm ...... You're trying to get those moments by first finding the moment generating function ......? Big mistake. Unless you totally want to work with erf(x) ...... (I'm betting you don't).
So how to get those pesky moments ......
You've gotta go back to first principles, sport. And it's your lucky day because the hard work's been done for you already ....
Read my replies in this thread
And a word to the wise ..... Next time post the real question .....