How do I show this :$\int_{-\infty}^{+\infty} x^n 2\cosh( x)e^{-x^2}=0$ if it is true with $n$ odd positive integer?

Question

I have looked to show that this integral:

$$\int_{-\infty}^{+\infty} x^n 2\cosh( x)e^{-x^2}=0$$

for $n$ is an odd positive integer , but i don't succeed to show that using standard method for getting closed form , Wolfram alpha show that is $0$ for some odd positive integer as shown here for $n=3$ , then my question here is :

Question : How do I show that integral is $0$ for odd positive integer $n$ if it is true ?

Answer 1

The function is odd at this case and makes the integral zero(because of strong exponential term in function equation the integral converges)

Answer 2

The integrand is an odd function in that case, being the product of an odd function, a constant, and two even functions. The integral of an odd function over a symmetric domain is $0$.

Answer 3

The function $f_n$ defined by $$ f_n\left(x\right)=2x^n\text{cosh}\left(x\right)e^{-x^2} $$ is odd when $n=2p+1$ ($-f_n\left(x\right)=f_n\left(-x\right)$. Hence $$ \int_{0}^{+\infty}f_n\left(x\right)\text{d}x=-\int_{-\infty}^{0}f_n\left(x\right)\text{d}x $$

Answer 4

Hint : $$\int_{-a}^a f(x) \rm{d} x =0$$ if $f(x)$ is an odd function.