Answer to Question #91396 in Calculus for Sajid

Question

Q. Solve ∫_(-∞)^∞▒x^2 e^(-x^2 )cos xdx.

Answer 1

$$cos(x)=\sum_{n=0}^\infin \frac{(-1)^n x^{2n}}{(2n)!}$$

so

$$I = \int_{-\infin}^{\infin} x^2 e^{-x^2} cos(x)dx=$$ $$= \int_{-\infin}^{\infin} x^2 e^{-x^2} \sum_{n=0}^\infin \frac{(-1)^n x^{2n}}{(2n)!}dx=$$

$$=\sum_{n=0}^\infin \frac{(-1)^n}{(2n)!}\int_{-\infin}^\infin e^{-x^2} x^{2n+2} dx=$$

$$=\sum_{n=0}^\infin \frac{(-1)^n}{(2n)!} \;2\int_{0}^\infin e^{-x^2} x^{2n+2} dx=|x \mapsto \sqrt{x}|=$$

$$=\sum_{n=0}^\infin \frac{(-1)^n}{(2n)!}\;2*1/2 \int_{0}^\infin e^{-x} x^{n+1-1/2} dx=$$ $$=\sum_{n=0}^\infin \frac{(-1)^n}{(2n)!} \int_{0}^\infin e^{-x} x^{n+1/2} dx=$$

$$=\sum_{n=0}^\infin \frac{(-1)^n}{(2n)!}\Gamma(n+3/2)=$$

$$=\sum_{n=0}^\infin \frac{(-1)^n}{(2n)!}(n+1/2) \Gamma(n+1/2)=$$ $$=\sum_{n=0}^\infin \frac{(-1)^n}{(2n)!} \frac{(n+1/2)(2n)!}{4^n n!} \sqrt\pi=$$

$$=\sum_{n=0}^\infin \frac{(-1)^n(n+1/2)}{4^n n!}\sqrt\pi=$$

$$=1/2\sqrt\pi\sum_{n=0}^\infin\frac{(-1/4)^n}{n!}+\sqrt\pi\sum_{n=0}^\infin\frac{(-1)^n n}{4^n n!}=$$

$$=1/2\sqrt\pi\sum_{n=0}^\infin\frac{(-1/4)^n}{n!}-1/4\sqrt\pi\sum_{n=0}^\infin\frac{(-1)^n }{4^n n!}=$$

$$=1/4\sqrt\pi e^{-1/4}=\frac{\sqrt\pi}{4\sqrt[4]{e}}.$$