Integral of $\int_0^{\infty} x^{4n+3} e^{-x} \sin x dx$.

Question

Can some one help me with the integral $$\int_0^{\infty} x^{4n+3} e^{-x} \sin x dx$$ According to my exercise I should be able to get $0$. Please help me .

Answer 1

Write the integral as the imaginary part of

$$\int_0^{\infty} dx \, x^{4 n+3} \, e^{-(1-i) x} $$

(I now assume $n$ is a positive integer.) Sub $u=(1-i) x$ and get

$$(1-i)^{-(4 n+4)} \int_0^{\infty} du \, u^{4 n+3} \, e^{-u} $$

(NB The integral is really along a ray in the complex plane, but this is not important as the integrand is analytic in the complex plane.)

Now,

$$(1-i)^{-(4 n+4)} = \frac1{4^{n+1}} e^{-i (n+1)\pi} = \left ( -\frac14\right)^{n+1} $$

which is obviously real. As the imaginary part of a real quantity is zero, ...