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I understand how to do the computation here:

$$\int_{0}^{\infty}xe^{-x^2}dx = \frac{1}{2}\int_{-\infty}^{0}e^{u}du = \frac{1}{2}$$

But I don't have any intuition for why the right-hand area under the curve $y=xe^{-x^2}$ would happen to be exactly half the left-hand area under the curve $y=e^x$. Or more generally, why this relationship holds:

$$\int_a^b xe^{-x^2}dx = \frac12\int_{-b^2}^{-a^2} e^xdx$$

Just to be clear, I'm not at all confused about the computation itself. I get that $u=-x^2$ and then you just substitute all the right things in. What I'm looking for is something like a geometric or behavioral intuition for why these two functions ($x \mapsto xe^{-x^2}$ and $x \mapsto e^{x}$) have such closely related areas under the curve.

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  • $\begingroup$ The assertion $\int_a^b xe^{-x^2}\,dx=\frac{1}{2}\int_a^b e^x\,dx$ is wrong for most values of $a$ and $b$, the second integral goes from $a^2$ to $b^2$. For the modified assertion, one can give an area argument based on "local" scaling. $\endgroup$
    – André Nicolas
    Commented Mar 29, 2015 at 23:53
  • $\begingroup$ Ah, yes, I wasn't paying close enough attention to the bounds during the $u$-substitution since that translation doesn't play much of a role in the improper integral. I've corrected the error. $\endgroup$
    – Michael Smith
    Commented Mar 30, 2015 at 0:31
  • $\begingroup$ The substitution rule for the antiderivative is the chain rule backwards, and the chain rule has easy intuitive explanation if we think of derivative as linear approximation. For the definite integral, which is the more interesting issue, Riemann sums for $xe^{-x^2}$ are closely related to Riemann sums for one transfer to Riemann sums for the other. $\endgroup$
    – André Nicolas
    Commented Mar 30, 2015 at 4:27

1 Answer 1

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This is just integration by substitution,

Let $u=-x^2$, then $du=-2x dx$ and $\frac{-1}{2}du=x dx$

$$$$

$$ \int^b_a xe^{-x^2}dx = -\frac{1}{2}\int e^{u} du $$

Now, $x=b \iff u=-b^2$ and $x=a \iff u=-a^2$, so the integral becomes

$$\begin{align}\int^b_a xe^{-x^2}dx &= -\frac{1}{2}\int^{-b^2}_{-a^2} e^{u} du\\ &= \frac{1}{2}\int_{-b^2}^{-a^2} e^{u}du \end{align}$$

To get the intuition, consider,

$$\int_a^b xe^{-x^2}dx = \int_a^b \frac{d}{dx}\left(\frac{-1}{2}e^{-x^2} \right) dx= \frac{-1}{2}e^{-x^2}|_a^b=\frac{-1}{2}(e^{-b^2}-e^{-a^2})$$

and $$\int_c^d e^xdx= e^x|_c^d=e^d-e^c$$

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    $\begingroup$ There's a factor of 2 missing in there somewhere. $\endgroup$
    – Deepak
    Commented Mar 29, 2015 at 23:14
  • $\begingroup$ Thanks @Deepak. Got too excited $\endgroup$
    – Trajan
    Commented Mar 29, 2015 at 23:15
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    $\begingroup$ Wait this isn't right, the derivative needs to be inside the integrand for this to make any sense. $\endgroup$
    – JLA
    Commented Mar 30, 2015 at 0:41
  • $\begingroup$ @JLA The derivative can be taken outside of the integral, in pure mathematics this would have to be rigorously justified, but in applied mathematics you can often just ignore the details $\endgroup$
    – Trajan
    Commented Jan 1, 2016 at 15:55
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    $\begingroup$ I don't think OP is confused about the substitution, but he is rather looking for an explanation about the relationship between the two functions. $\endgroup$
    – rubik
    Commented Jan 2, 2016 at 12:45

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