Conjecture: $n\in\mathbb{Z}^+\Rightarrow 0=\int_{0}^\infty x^{4n-1}\sin(x)e^{-x}dx$

Question

Conjecture: $n\in\mathbb{Z}^+\Rightarrow 0=\int_{0}^\infty x^{4n-1}\sin(x)e^{-x}dx$

This has been verified with WolframCloud for $1\leq n\leq 500$.

Answer 1

Consider $$ c_{n} = \int_{0}^{\infty} x^{n} e^{(-(1-i)x)}dx = a_{n} + ib_{n} $$ where $$ b_{n} = \int_{0}^{\infty} x^{n} \sin x e^{-x} dx. $$ By using the substitution $y = (1-i)x$, we get $$ c_{n} = \frac{1}{(1-i)^{n+1}} \int_{0}^{\infty} y^{n} e^{-y} dy = \frac{n!e^{(n+1)i\pi/4}}{2^{n/2}} = \frac{n!}{2^{n/2}}\left[ \cos \frac{(n+1)\pi}{4} + i\sin \frac{(n+1)\pi}{4}\right]. $$ From $\Gamma(n+1) = n!$. Hence, $$ b_{n} = \frac{n!}{2^{n/2}} \sin \frac{(n+1)\pi}{4} $$ which shows $b_{4n-1} = 0$.

Answer 2

Assuming $x >0$, consider the more general problem of $$I=\int x^m e^{-x}e^{ix}\,dx=\int x^m e^{-(1-i) x} \,dx=-x^{m+1} E_{-m}((1-i) x)$$ $$J=\int_0^\infty x^m e^{-x}e^{ix}\,dx=(1-i)^{-(m+1)} \Gamma (m+1)$$ Now, make $m=4n-1$ to get $$J=4^{-n} e^{i \pi n} \Gamma (4 n)$$ Consider the imaginary part $$\Im(J)=\Im(e^{i \pi n})4^{-n} \Gamma (4 n)=4^{-n} \Gamma (4 n)\sin(n \pi)$$ So, if $n$ is an integer, your result.