Why is $\int_{-∞}^∞ x^k*e^x/(e^x +1)^2 dx = 0$ for odd k?

Question

Here's what I've considered:

• $e^x$ and $(e^x + 1)^2$ are of no parity (neither odd nor even functions).
• $e^k$ for odd k is an odd function.

I've always thought one could only multiply functions of some parity, but according to this result (from Salinas, Introduction to Statistical Physics) it is also possible to say

• odd function * no parity function = odd function

So my two questions are:

1. Why is this correct?
2. Is it right to draw the same conclusion for even function * no parity function?

Thank you

Answer 1

$$\dfrac{e^x}{(e^x+1)^2} = \dfrac{1}{(e^{x/2} + e^{-x/2})^2}$$ is an even function.