My Take on Mathematical Reflection Problem

This is problem U610 from Mathematical Reflection, Edition 6 2022.

If n is a positive integer, evaluate

\int_0^{+\infty} x^{n-1}e^{-x} \left(\sum_{k=0}^n \binom{n}{k} \left(-1 \right)^k \cos \left(x+ \frac{k \pi}{2}\right)\right)dx.

Solution : Write

C=\int_0^{+\infty} x^{n-1}e^{-x} \left(\sum_{k=0}^n \binom{n}{k} \left(-1 \right)^k \cos \left(x+ \frac{k \pi}{2}\right)\right)dx, S = \int_0^{+\infty} x^{n-1}e^{-x} \left(\sum_{k=0}^n \binom{n}{k} \left(-1 \right)^k \sin \left(x+ \frac{k \pi}{2}\right)\right)dx,

we want to evaluate the value of C = Re\left(C+i.S\right), while on the other hand

C+i.S = \int_0^{+\infty} x^{n-1}e^{-x} \left(\sum_{k=0}^n \binom{n}{k} \left(-1 \right)^k \cos \left(x+ \frac{k \pi}{2}\right)+i.\sin \left(x+ \frac{k \pi}{2}\right)\right)dx

= \int_0^{+\infty} x^{n-1}e^{-x} e^{i.\left(x+\frac{k \pi}{2}\right)}dx

= \int_0^{+\infty} \left(1-i\right)^n x^{n-1} e^x e^{-x} dx

= \left(1-i\right) \int_0^{+\infty} \left(\left(1-i\right)x\right)^{n-1} e^{\left(-1+i\right)x} dx.

Take u = \left(1-i\right)x, we thus have du = \left(1-i\right) dx, and

\left(1-i\right) \int_0^{+\infty} \left(\left(1-i\right)x\right)^{n-1} e^{\left(-1+i\right)x} dx = \left(1-i\right) \int_0^{+\infty} \left(\left(1-i\right)x\right)^{n-1} e^{\left(-1+i\right)x} dx=\int_0^{+\infty} u^{n-1} e^u du = \Gamma(n) = (n-1)! = C.

Corollary : If n is a positive integer, then we have

\int_0^{+\infty} x^{n-1}e^{-x} \left(\sum_{k=0}^n \binom{n}{k} \left(-1 \right)^k \sin \left(x+ \frac{k \pi}{2}\right)\right)dx = 0