Show that $\int x\mathrm{e}^{-\alpha x^2}\mathrm dx =\dfrac{-1}{2\alpha} \mathrm e^{-\alpha x^2}$ + Constant I tried to do this integration by parts and got  $\int x\mathrm{e}^{-\alpha x^2}\mathrm dx =\dfrac{-1}{2\alpha} \mathrm e^{-\alpha x^2} +\alpha\int x^3\mathrm{e}^{-\alpha x^2}\mathrm dx$ + constant. Where $\alpha$ is a constant.
Any help will be most appreciated.
Thank you.
 A: Hint: Put $-\alpha x^2=u \Rightarrow -2\alpha x \mathrm dx=\mathrm du$,then $x\mathrm dx=\frac {-1}{2\alpha}\mathrm du$. you will have,
$$\displaystyle\int xe^{-\alpha x^2}\,dx=\frac {-1}{2\alpha}\displaystyle\int e^u \mathrm du$$
A: Hint: I don't think integration by parts is the right strategy (and I don't understand the way you used it). Try performing the substitution $u=x^2$, $du = 2xdx$.
A: Find the derivative of $x\mapsto e^{-\alpha x^2}$. You'll see that it's easy to find an antiderivative of $x\mapsto xe^{-\alpha x^2}$ and thus to solve this integral, which by the way is not a function of $x$.
A: If $\alpha=0$ the integral diverges (it is $\infty$). Now if $\alpha \not =0$, the derivative of $\frac{-1}{2\alpha}e^{-\alpha x^{2}}$ is $xe^{-\alpha x^{2}}$. Now $\int\limits_{0}^{\infty} xe^{-\alpha x^{2}}dx=\frac{1}{2\alpha}$. The answer should be a real number, and not a function of $x$
A: Are you familiar with integrals of the form $\int f'(x).g(f(x)) \text{d}x$, where $f$ and $g$ are ''well-behaved'' functions? If yes, try to see how you can apply this idea when $g(x)=e^x$ and $f(x)=\alpha x^2$, where $\alpha$ is a constant of course. 
If not, find out how to use this method because it is an essential tool. Actually, knowing about differentiation and the idea behind integration, you could try finding out what happens yourself (which is why I did not give out the result of the integration).
