Integral: $\int_{-\infty}^{\infty} x^2 e^{-x^2}\mathrm dx$ 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.
 A: 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:
$$
  \int_{-\infty}^\infty x^2 \mathrm{e}^{-x^2} \mathrm{d} x = \left.  -\frac{\mathrm{d}}{\mathrm{d} s} \sqrt{\frac{\pi}{s}} \right|_{s=1} = \left. \frac{\sqrt{\pi}}{2} s^{-\frac{3}{2}} \right|_{s=1} = \frac{\sqrt{\pi}}{2}
$$
A: (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. 
A: 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
$$
I=\frac{\sqrt{\pi }}{2}.\tag{6}
$$
A: 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.)
A: 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,
$$\frac{d^2}{dx^2}e^{-x^2}=-2e^{-x^2}+4x^2e^{-x^2}$$
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 
$$\begin{array}{rcl}
0&=&\int -2e^{-x^2}+4x^2e^{-x^2}\,dx\\
&=&-2\sqrt\pi+4\int x^2e^{-x^2}\,dx\\
\implies\int x^2e^{-x^2}\,dx&=&\frac{\sqrt\pi}{2}
\end{array}
$$
