Question

Integrate: y=(x^2)e^(-pi(x^2)) bounded by minus infinity and infinity

Answer 1

\[\int x^2e^{-\pi x^2}dx\]
\[\begin{matrix}u_1=x& & &dv_1=xe^{-\pi x^2}dx\\
du_1=dx& & &v_1=-\frac{1}{2\pi}e^{-\pi x^2}\end{matrix}\]
\[\int x^2e^{-\pi x^2}dx=-\frac{1}{2\pi}xe^{-\pi x^2}+\frac{1}{2\pi}\int xe^{-\pi x^2}dx\]
\[t=-\pi x^2\\
dt=-2\pi x\;dx\]
\[\begin{align*}\int x^2e^{-\pi x^2}dx&=-\frac{1}{2\pi}xe^{-\pi x^2}+\frac{1}{2\pi}\int e^{t}dt\\
&=-\frac{1}{2\pi}xe^{-\pi x^2}+\frac{1}{2\pi}e^t+C\\
&=-\frac{1}{2\pi}xe^{-\pi x^2}+\frac{1}{2\pi}e^{-\pi x^2}+C\end{align*} \]
Evaluated over (-∞, ∞):
\[\begin{align*}\int_{-\infty}^{\infty} x^2e^{-\pi x^2}dx&=\int_{-\infty}^{0} x^2e^{-\pi x^2}dx+\int_{0}^{\infty} x^2e^{-\pi x^2}dx\\
&=\lim_{a\to-\infty}\int_{a}^{0} x^2e^{-\pi x^2}dx+\lim_{b\to\infty}\int_{0}^{b} x^2e^{-\pi x^2}dx\\
&=\lim_{a\to-\infty}\left[-\frac{1}{2\pi}xe^{-\pi x^2}+\frac{1}{2\pi}e^{-\pi x^2}\right]_a^0\\&\;\;\;\;\;\;\;\;\;+\lim_{b\to\infty}\left[-\frac{1}{2\pi}xe^{-\pi x^2}+\frac{1}{2\pi}e^{-\pi x^2}\right]_0^b\\
&=-\frac{1}{2\pi}\lim_{a\to-\infty}\left[xe^{-\pi x^2}-e^{-\pi x^2}\right]_a^0\\&\;\;\;\;\;\;\;\;\;-\frac{1}{2\pi}\lim_{b\to\infty}\left[xe^{-\pi x^2}-e^{-\pi x^2}\right]_0^b \end{align*}\]

Answer 2

I'll leave the limit evaluation to you

Answer 3

Oh, and ignore the subscripts for u and dv in the beginning. I expected another round of IBP, but it never happened and I forgot to take them out.