I don't know how to evaluate it. I know there is one method using the gamma function. BUT I want to know the solution using a calculus method like polar coordinates.

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

I will wait for a solution. Thank you.

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    $\begingroup$ Here is a more general question. $\endgroup$ Commented Sep 22, 2011 at 12:11
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    $\begingroup$ The question "How do I find $\int_{-\infty}^\infty e^{-x^2}\,dx$" has been asked and answered on this forum many times. I'm surprised so many people bothered to take the time to type a solution as part of the answer to your question when it has been done before. You can convert your integral to that one using integration by parts, with $dv = xe^{-x^2}dx$ and $u = x$. Also, J.M.'s link answers your question. $\endgroup$ Commented Oct 18, 2013 at 22:56

5 Answers 5


In order to solve the integral by polar coordinates first consider $I_s = \int_{-\infty}^\infty \mathrm{e}^{-s x^2} \mathrm{d} x$. The integral you seek will be obtained by differentiation as $-\left. \frac{\mathrm{d}}{\mathrm{d} s} I_s \right|_{s=1}$.

Now, to evaluate $I_s$:

$$ I_s^2 = \int_{-\infty}^\infty \mathrm{e}^{-s x^2} \mathrm{d} x \cdot \int_{-\infty}^\infty \mathrm{e}^{-s y^2} \mathrm{d} y = \int_{-\infty}^\infty \int_{-\infty}^\infty \mathrm{e}^{-s (x^2 + y^2)} \, \mathrm{d} x \mathrm{d} y $$ Now change variables into polar coordinates $x = r \sin \theta$ and $y = r \cos \theta$. $$ I_s^2 = \int_{0}^{2 \pi} \mathrm{d} \theta \int_0^\infty \mathrm{e}^{-s r^2} \cdot r \, \mathrm{d} r = \pi \int_0^\infty \mathrm{e}^{-s t} \mathrm{d} t = \frac{\pi}{s} $$ where $t = r^2$ change of variable has been made.

Now, since $I_s > 0$ for $s >0$, we obtain $I_s = \sqrt{\frac{\pi}{s}}$.

The integral in question now follows: $$ \int_{-\infty}^\infty x^2 \mathrm{e}^{-x^2} \mathrm{d} x = \left. -\frac{\mathrm{d}}{\mathrm{d} s} \sqrt{\frac{\pi}{s}} \right|_{s=1} = \left. \frac{\sqrt{\pi}}{2} s^{-\frac{3}{2}} \right|_{s=1} = \frac{\sqrt{\pi}}{2} $$

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    $\begingroup$ Very nice indeed! A solution that yields the value of $\int_{-\infty}^{\infty} x^{2n} e^{-x^2} \mathrm dx$ upon differentiating $n$ times with respect to $s$ and setting $s = 1$... $\endgroup$ Commented Sep 20, 2011 at 15:21
  • $\begingroup$ Forgive me if this is obvious, but why are we setting (or how do we know) s = 1? $\endgroup$ Commented Jan 18 at 20:11

(As the OP wants a solution without using the gamma function.) Following Davide's suggestion, we write: $$ \int_{-\infty}^{\infty} x^2 e^{-x^2} dx = -\frac{1}{2}\int_{-\infty}^{\infty} x \cdot (-2x e^{-x^2}) dx. $$ Let $u = x$ and $v = e^{-x^2}$. We have $\frac{dv}{dx} = -2xe^{-x^2}$. Integrating by parts: $$ -\frac{1}{2}\int_{-\infty}^{\infty} u \frac{dv}{dx} dx = -\frac{1}{2} \left. uv \right|_{-\infty}^{\infty} + \frac{1}{2} \int_{-\infty}^{\infty} v \frac{du}{dx} dx. $$ I will leave it as an exercise to compute the first term. (Hint: it should come out to $0$. :)) The integral appearing in the second term (ignoring the factor of $\frac{1}{2}$ in the front) simplifies to: $$ \int_{-\infty}^{\infty} v \frac{du}{dx} dx = \int_{-\infty}^{\infty} e^{-x^2} dx. $$ This is the famous Gaussian integral, whose value is $\sqrt{\pi}$. You should now be able to evaluate integral easily.

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    $\begingroup$ [For those who had trouble solving hint] First part can be reduced to $\lim_{a\to 0} a\sqrt{ln(1/a)}$ by substituting $a = e^{-x^{2}}$ which can be further reduced to $\lim_{a\to 0^{+}} \sqrt{ln(1/a^{2})/2*a^{2}}$ and since we know that $\lim_{x\to 0} ln(1/x)/x = 0$, the previous also evaluates to zero. $\endgroup$ Commented Feb 14, 2017 at 20:35
  • $\begingroup$ [For those who had trouble solving hint] an alternative way of evaluating the term which in intended to be zero, is to acknowledge that if $f(x) = xe^{-x^2}$ then $f(-x) = (-x)e^{-(-x)^2} = -f(x)$, or in other words, the term represents a integral over an symmetric interval, and thus, its 0. $\endgroup$
    – JustDanyul
    Commented Mar 13, 2019 at 20:26
  • $\begingroup$ @JustDanyul I think that's not correct, since inside the vertical bar f(x) isn't approaching +inf and -inf at the same speed. Actually if f(x) diverges approaching +inf and -inf, the result would diverge. The same reason why $$x^3|^{\infty}_{-\infty}$$ diverges. $\endgroup$
    – Hot.PxL
    Commented May 4, 2020 at 12:43

Detailing Srivatsan Narayanan's solution. It is known that the functional equation of the gamma function may be derived applying the integration by parts technique. Its value at $1/2$ may be evaluated by computing a double integral over the first quadrant in Cartesian and polar coordinates. Let's apply similar ideas in this case. Let $f(x)=x^{2}e^{-x^{2}}$. Since $ f(-x)=f(x)$ the integral $\int_{-\infty }^{\infty }f(x)\mathrm{d}x=2\int_{0}^{\infty }f(x)\mathrm{d}x$. Integrating by parts $$ \int x^{2}e^{-x^{2}}\mathrm{d}x=\int x\cdot xe^{-x^{2}}\mathrm{d}x, $$ since $$ \int xe^{-x^{2}}\mathrm{d}x=-\frac{1}{2}e^{-x^{2}} $$ and $\frac{dx}{dx}=1$, we get $$\begin{eqnarray*} \int x^{2}e^{-x^{2}}\mathrm{d}x &=&x\left( -\frac{1}{2}e^{-x^{2}}\right) -\int -\frac{ 1}{2}e^{-x^{2}}\,\mathrm{d}x \\ &=&-\frac{1}{2}xe^{-x^{2}}+\frac{1}{2}\int e^{-x^{2}}\,\mathrm{d}x.\tag{0} \end{eqnarray*}$$

And so, $$\begin{eqnarray*} \int_{0}^{\infty }x^{2}e^{-x^{2}}dx &=&\left. -\frac{1}{2} xe^{-x^{2}}\right\vert _{0}^{\infty }+\frac{1}{2}\int_{0}^{\infty }e^{-x^{2}}\,\mathrm{d}x \\ &=&\left( \lim_{c\rightarrow \infty }-\frac{1}{2}ce^{-c^{2}}\right) +\frac{1 }{2}0e^{-0^{2}}+\frac{1}{2}\int_{0}^{\infty }e^{-x^{2}}\,\mathrm{d}x \\ &=&0+0+\frac{1}{2}\int_{0}^{\infty }e^{-x^{2}}\,\mathrm{d}x \\ &=&\frac{1}{2}\int_{0}^{\infty }e^{-x^{2}}\,\mathrm{d}x.\tag{1} \end{eqnarray*}$$

Consequently, $$ I:=\int_{-\infty }^{\infty }x^{2}e^{-x^{2}}\mathrm{d}x=2\int_{0}^{\infty }x^{2}e^{-x^{2}}\mathrm{d}x=\int_{0}^{\infty }e^{-x^{2}}\,\mathrm{d}x.\tag{2} $$

To evaluate this last integral we compute the following double integral in Cartesian and polar coordinates ($r^{2}=x^{2}+y^{2}$, $x=r\cos \theta ,y=r\sin \theta $). Since the Jacobian of the transformation $\frac{\partial (x,y)}{\partial (r,\theta )}=r$, we have $$ \int_{x=0}^{\infty }\int_{y=0}^{\infty }e^{-x^{2}-y^{2}}\mathrm{d}x\mathrm{d}y=\int_{\theta =0}^{\pi /2}\int_{r=0}^{\infty }e^{-r^{2}}r\mathrm{d}r\mathrm{d}\theta.\tag{3} $$ Comparing the LHS $$ \int_{x=0}^{\infty }\int_{y=0}^{\infty }e^{-x^{2}-y^{2}}\mathrm{d}x\mathrm{d}y=\left( \int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x\right) \left( \int_{0}^{\infty }e^{-y^{2}}\mathrm{d}y\right) =I^{2}\tag{4} $$ with the RHS $$\begin{eqnarray*} I^2 &=&\int_{\theta =0}^{\pi /2}\int_{r=0}^{\infty }e^{-r^{2}}r\mathrm{d}r\mathrm{d}\theta = \frac{\pi }{2}\int_{0}^{\infty }e^{-r^{2}}r\mathrm{d}r \\ &=&\frac{\pi }{2}\left. \left( -\frac{1}{2}e^{-r^{2}}\right) \right\vert _{0}^{\infty }=\frac{\pi }{2}\left( \lim_{c\rightarrow \infty }-\frac{1}{2} e^{-c^{2}}+\frac{1}{2}e^{-0^{2}}\right) \\ &=&\frac{\pi }{2}\left( 0+\frac{1}{2}\right) =\frac{\pi }{4},\tag{5} \end{eqnarray*}$$ yields $$ I=\frac{\sqrt{\pi }}{2}.\tag{6} $$


First, since the integrand is symmetric around $0$, we can write it as twice the integral from $0$ to $\infty$. Now, change variables by letting $u=x^2$ so that $du=2xdx$. Then our integral becomes $$\int_{-\infty}^\infty x^2e^{-x^2} dx=\int_{0}^\infty xe^{-x^2} 2xdx=\int_{0}^\infty u^{\frac{1}{2}} e^{-u}du=\Gamma\left(\frac{3}{2}\right) =\frac{\sqrt{\pi}}{2}$$ by the definition of the Gamma function along with the fact that $\Gamma(1/2)=\sqrt{\pi}$. (7 proofs of this last identity, or equivalently the identity $\int_{-\infty}^\infty e^{-x^2}dx =\sqrt{\pi}$ are given on this Math Stack Exchange post.)

  • $\begingroup$ Thank you. But I want to know solution not using gamma function. $\endgroup$ Commented Sep 20, 2011 at 14:40
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    $\begingroup$ @Kim: In any case, you will need to use the fact that $\int_{-\infty}^\infty e^{-x^2}dx =\sqrt{\pi}$. This integral by a change of variables is the same as $\Gamma(1/2)$. Any solution will implicitly be using or proving facts about the Gamma function. $\endgroup$ Commented Sep 20, 2011 at 14:42

If you know that $\int_{-\infty}^\infty e^{-x^2}=\sqrt\pi$, then you can use it to easily solve this.

Differentiating $e^{-x^2}$ twice,


Now integral of the left hand side is $0$, as $\int\left(\frac{d^2}{dx^2}e^{-x^2}\right)dx=\frac{d^2}{dx^2}\int e^{-x^2}dx=\frac{d^2}{dx^2}\sqrt\pi=0$.

So integrating, we get

$$\begin{array}{rcl} 0&=&\int -2e^{-x^2}+4x^2e^{-x^2}\,dx\\ &=&-2\sqrt\pi+4\int x^2e^{-x^2}\,dx\\ \implies\int x^2e^{-x^2}\,dx&=&\frac{\sqrt\pi}{2} \end{array} $$


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