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What is the result of the following expression?

\begin{equation} \int^{\infty}_{-\infty}dx\;x\;e^{-x^2} \end{equation}

On trying to use $u=x^2$ and evaluate the above, I get $\frac{\sqrt \pi}{2}$, but my homework solution says it is $0$ as $x\;e^{-x^2}$ is an odd function. How is this possible?

EDIT (my steps):

Setting $u=x^2$, $du = 2xdx$. Using this in the integral results in,

\begin{equation} \int^{\infty}_{-\infty}\frac{du}{2}e^{-u^2} \end{equation}

Taking $\frac{1}{2}$ out of the integral, the rest is $\int^{\infty}_{-\infty} e^{-u^2} = \sqrt\pi$ and so the result $\frac{\sqrt\pi}{2}$.

What is wrong with the approach above?

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    $\begingroup$ You'd have to show us the different steps in your evaluation of the integral for us to find if you made a possibly tiny mistake in the calculations. By the way, it might be safer tu use the subsitution $u=-x^2$. $\endgroup$
    – MasB
    Oct 25, 2020 at 11:49
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    $\begingroup$ actually you cannot substitute $u = x^2$ as the function is not invertible in the given range $\endgroup$ Oct 25, 2020 at 12:00
  • $\begingroup$ with $u=x^2$ the extremes of integration becom both $+\infty$ and so the integral should be zero, shouldn't it? $\endgroup$
    – Raffaele
    Oct 25, 2020 at 13:04

1 Answer 1

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You have two mistakes in your substitution.

You substitute $u = x^2,$ but you write $e^{-u^2}$ after replacing $x^2$ in $e^{-x^2}.$ (This is technically an error but I think it actually is not responsible for the error in the answer.)

The critical error is that after substitution your bounds are $-\infty$ and $\infty.$ After the substitution $u=x^2,$ there are no negative values of $u.$ As $x$ increases from $-\infty$ to $0$, $u$ decreases from $\infty$ to $0$.

You can fully account for this by writing your substitution in two parts:

\begin{align} \int_{-\infty}^\infty dx\,xe^{-x^2} &= \int_{-\infty}^0 dx\,xe^{-x^2} + \int_0^\infty dx\,xe^{-x^2} \\ &= \int_{\infty}^0 du\,\frac12 e^{-u} + \int_0^\infty du\,\frac12 e^{-u}. \\ \end{align}

Can you finish it from there?

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  • $\begingroup$ Thanks David K. I see my folly in substitution - the integrand becomes $𝑒^{−𝑢}$ on substitution. And it is much more clear when the limits are split you put, giving (-1 -0) + (-0+1). $\endgroup$
    – RK1974
    Oct 25, 2020 at 18:45

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