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I was working on the following definite integral:

$\displaystyle\int^{\infty}_{-\infty}{xe^{-x^{2}}}dx$

I decided to use $u$-substitution to make the integral simpler, by letting $u = -x^{2} \Rightarrow du = - 2x\hspace{1 mm}dx$.

The lower bound then becomes $-(-\infty)^{2}=-\infty$, while the upper bound becomes $-\infty^{2} = -\infty$.

So the overall definite integral is then:

$= -\dfrac{1}{2}\displaystyle\int^{-\infty}_{\infty}{e^{u}}du$

Which evaluates to:

$= -\dfrac{1}{2} \bigg(e^{-\infty} - e^{-\infty}\bigg)$

$= -\dfrac{1}{2} \times 0$

$= 0$

Is this a legitimate way of doing this problem? Or were there any errors made?

Thank you.

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2 Answers 2

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Your answer is correct but the method is sloppy.

Note that $$ \displaystyle\int^{\infty}_{-\infty}{xe^{-x^{2}}}dx$$ is defined as $$ \displaystyle\int^{0}_{-\infty}{xe^{-x^{2}}}dx +\displaystyle\int^{\infty}_{0}{xe^{-x^{2}}}dx $$

Now you apply your substitution and get the same result.

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In fact you can simply use that for $N>0$ $$ \int_{-N}^{N}xe^{-x^2}\text{d}x=-\frac{1}{2}\left(e^{-N^2}-e^{-N^2}\right)=0 $$ Then taking $\displaystyle N \rightarrow +\infty$ leads you to the result.

I think it is a better approach as $e^{\pm \infty}$ has no sense ( for me )

Note that $\displaystyle \left(-\frac{1}{2}e^{-x^2}\right)'=xe^{-x^2}$.

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    $\begingroup$ In general, this answer is not always applicable. For example, we cannot use the same argument to say that $\int_{-\infty}^{\infty} xdx = 0$ because both 'halves of the integral diverge. I would try breaking your problem up into two integrals: one from $-\infty$ to $0$, and one from $0$ to $\infty$. $\endgroup$
    – D.B.
    May 13, 2018 at 15:06

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