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How do I evaluate this integral using simple calculus techniques such as substitution or parts?

$$\int\limits_{-\infty}^{\infty}(1-x^{2})e^{-x^{2}}\mathrm{d}x$$

I would not like any solution involving more advanced techniques such as the gamma function, the error function or polar coordinates.

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closed as off-topic by Jack D'Aurizio, Daniel W. Farlow, user1551, heropup, user223391 Aug 28 '16 at 13:53

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  • $\begingroup$ Well, the part that's $e^{-x^2}$ you can do using the squaring and polar coordination trick. Not sure the other half $\endgroup$ – Alan Aug 25 '16 at 22:38
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    $\begingroup$ For the second piece you can integrate by parts to reduce it to the first case. $\endgroup$ – Ian Aug 25 '16 at 22:55
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    $\begingroup$ If you disallow polar coordinates, the error function and the $\Gamma$ function, you leave us without options. Where to get $\sqrt{\pi}$ from? Can we at least use $\int_{-\infty}^{+\infty}e^{-x^2}\,dx=\sqrt{\pi}$? $\endgroup$ – Jack D'Aurizio Aug 25 '16 at 23:01
  • $\begingroup$ @JackD'Aurizio Okay, fair enough. I was just looking for solutions that avoided those techniques where possible. $\endgroup$ – Cascadex Aug 25 '16 at 23:11
  • $\begingroup$ He didn't rule out complex analysis, right? mathoverflow.net/a/105462/454 $\endgroup$ – GEdgar Aug 25 '16 at 23:34
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Integrate by parts $$\int_{-\infty}^{\infty} e^{-x^2} \; dx = \left. xe^{-x^2}\right|_{-\infty}^{\infty}-\int_{-\infty}^{\infty} -2x^2 e^{-x^2} \; dx.$$ Using the classic "switch to polar" trick for the first integral gives $$\sqrt{\pi} = 2\int_{-\infty}^{\infty} x^2 e^{-x^2} \; dx.$$ So your integral equals $\sqrt{\pi} - \sqrt{\pi}/2 = \sqrt{\pi}/2.$

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  • $\begingroup$ And this shows that if you could do all this without the "polar trick", then you could integrate $e^{-x^2}$ without it as well, and no one seems to do that...suggesting that your question doesn't have a (known) answer. $\endgroup$ – John Hughes Aug 25 '16 at 23:19
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Here, we use "Feynman's Trick" for Differentiating Under the Integral as a straightforward way to evaluate integrals of the form $\int_{-\infty}^\infty x^{2n}e^{-x^2}\,dx$. To that end, we proceed.

Let $I(a)$ be the integral defined by

$$I(a)=\int_{-\infty}^\infty e^{-ax^2}\,dx \tag 1$$

The integral in $(1)$ is a Gaussian integral, which can be evaluated a number of ways (SEE HERE for a way that avoids using polar coordinates or the Gammma function ) to obtain

$$I(a)=\sqrt{\frac{\pi}a} \tag 2$$

Now, differentiating the right-hand side of $(1)$ reveals

$$I'(a)=-\int_{-\infty}^{\infty}x^2e^{-ax^2}\,dx \tag 3$$

while differentiating the right-hand side of $(2)$ reveals

$$I'(a)=-\frac12 \sqrt{\pi}a^{-3/2} \tag 4$$

Upon equating $(3)$ and $(4)$ and setting $a=1$, we find

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty x^2e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}}$$

We can continue to differentiate $(1)$ to obtain

$$\begin{align} I^{(n)}(a)&=(-1)^n \int_{-\infty}^\infty x^{2n}e^{-ax^2}\,dx\\\\ &=(-1)^n\frac{(2n-1)!!}{2^n}a^{-(2n+1)/2} \end{align}$$

from which we arrive at

$$\int_{-\infty}^\infty x^{2n}e^{-ax^2}\,dx=\frac{\sqrt{\pi}\,(2n-1)!!}{2^n}\,a^{-(2n+1)/2} \tag 5$$

Setting $a=1$ in $(5)$ yields

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty x^{2n}e^{-x^2}\,dx=\frac{\sqrt{\pi}\,(2n-1)!!}{2^n} }$$

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$\int_{-\infty}^{\infty} e^{-x^2} dx - \int_{-\infty}^{\infty} x^2e^{-x^2} dx$

Lets tackel this in pieces

$\int_{-\infty}^{\infty} x^2e^{-x^2} dx$ integrate by parts

$u = dx, dv = xe^{-x^2} dx\\ du = dx, v = -\frac 12 e^{-x^2}$

$-\frac x2 e^{-x^2} |_{-\infty}^{\infty} + \frac 12 \int e^{-x^2} dx \\ -\frac x2 e^{-x^2} |_{-\infty}^{\infty} = 0$

And what is left pairs with the piece we had not yet addressed.

Since the fucntion is even, $\frac 12 \int_{-\infty}^{\infty} e^{-x^2} dx = \int_{0}^{\infty} e^{-x^2} dx$

You said you don't want polar coordinates, which is actually the easy way to solve this one.

Consider:

$F(t) = \int_0^{\infty} \frac {e^{-t^2(x^2+1)}}{1+x^2} dx\\ F'(t) = \int_0^{\infty} 2t e^{-t^2(x^2+1)} dx$

Note that we are differentiating with respect to t and now with respect to x.

$F'(t) = -2e^-t^2 \int_0^{\infty} te^{-(tx)^2} dx\\ u = tx\\ du = t dx\\ F'(t) = -2e^-t^2 \int_0^{\infty} e^{-u^2} du$

So... $F'(t) = e^-t^2 (-2I)$ where I is a constant (and I is also the number we are looking for.)

The fundamntal theorm of calucus says:

$\int_a^b F'(t) dt = F(b) - F(a)\\ F(b)- F(a) = -2I \int_b^a e^-t^2 dt $

Set $a = 0,$ and $b$ approaches infinity. $F(\infty) - F(0) = -2I\int_0^infty e^-t^2 dt = -2 I^2$

$F(0) = \int_0^{\infty} \frac {1}{1+x^2} dx$ Which integrates to an inverse tangent function and equals $\frac \pi 2$

$F(\infty) = \int_0^{\infty} 0 dx = 0$

$F(\infty) - F(0) = -\frac \pi 2 = -2 I^2\\ I^2 = \frac {\pi}4\\ I = \frac {\sqrt{\pi}}2$

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  • $\begingroup$ Do you mean $\sqrt{\pi}/2$? $\endgroup$ – Hrhm Aug 25 '16 at 23:42
  • $\begingroup$ You mistyped the upper limit in one of your integrals. You meant $\infty$ but left off the \ symbol, so the limit is $i$ and there are extra letters in the integrand. $\endgroup$ – Oscar Lanzi Aug 26 '16 at 0:09
  • $\begingroup$ Thanks, I had to shut down before I had a chance to check my work. $\endgroup$ – Doug M Aug 26 '16 at 16:10
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$$\int_{-\infty}^{\infty}\exp(-x^2)dx = \sqrt{\pi}$$

Substituting $x = \sqrt{a} t$ in here gives:

$$\int_{-\infty}^{\infty}\exp(-a t^2)dx = \sqrt{\frac{\pi}{a}}$$

Put $a = 1+u$ and expand both sides in powers of $u$. The integrand can be expanded as:

$$\exp\left[-(1+u)t^2\right]= \exp(-ut^2)\exp(-t^2)=\left[1-ut^2 +\mathcal{O}(u^2)\right]\exp(-t^2)$$

We also have:

$$\sqrt{\frac{\pi}{1+u}}=\sqrt{\pi}\left(1-\frac{1}{2}u+\mathcal{O}(u^2)\right)$$

Equating the integral to the result up to order $u$ yields:

$$\int_{-\infty}^{\infty}\left(1-ut^2\right)\exp(-t^2)dt =\sqrt{\pi}\left(1-\frac{1}{2}u\right)$$

The desired integral is thus $\frac{\sqrt{\pi}}{2}$.

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