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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.

BLAZE
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    The value of the integral $\displaystyle\int_{0}^{\infty}xe^{-\alpha x^2},dx$ should not be a function of $x$. – JimmyK4542 Jul 28 '15 at 07:09
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    You can't prove to be true that which is not true. The righthand side of the expression you want to prove is a function of $x$ but it can't be since the lefthand side is an integral on $x$ that leaves only $\alpha$ as a free variable. – wltrup Jul 28 '15 at 07:09
  • @JimmyK4542 Thanks for your reply I've changed it now – BLAZE Jul 28 '15 at 07:32
  • @wltrup Thanks for your reply I've changed it now – BLAZE Jul 28 '15 at 07:32

5 Answers5

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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,

$$\displaystyle\int xe^{-\alpha x^2}\,dx=\frac {-1}{2\alpha}\displaystyle\int e^u \mathrm du$$

Chiranjeev_Kumar
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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 $u=x^2$, $du = 2xdx$.

ajd
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    The result your lecturer gave is the indefinite integral. The definite integral is obtained by plugging $\infty$ and $0$ into the indefinite integral and subtracting. – ajd Jul 28 '15 at 07:26
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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 $x$.

Augustin
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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 $x$

mich95
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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).

BenBarrows
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