Show that $f(x)= x/(e^x -1)+x/2$ is even.

Question

Show that $f(x)= x/(e^x -1)+x/2$ is even.

So an even function is such that $f(-x)=f(x)$.

So I need to show that $f(x)= (-x)/(e^{-x} -1)+(-x)/2=x/(e^x -1)+x/2=f(x)$.

I also know that $\frac{x}{e^x-1}=\sum\limits_{n=0}^\infty \frac{B_nx^n}{n!}$ where $B_n$ are bernouli numbers.

Though I don't feel that gets me any closer as showing $\left(\sum\limits_{n=0}^\infty \frac{B_n(-x)^n}{n!}\right)+\frac{(-x)}{2}=\left(\sum\limits_{n=0}^\infty \frac{B_nx^n}{n!}\right)+\frac{x}{2}$

seams more difficult than showing $(-x)/(e^{-x} -1)+(-x)/2=x/(e^x -1)+x/2$

Is there a trick to this that I am not seeing?

Answer 1

You don't need to expand to power series, regular manipulations will do (using that $e^x\cdot e^{-x} = 1$). See below for a detailed derivation.

For any $x\in\mathbb{R}^\ast$, $$\begin{align} f(x)-f(-x) &= \frac{x}{e^x-1} + \frac{x}{2} - \left(\frac{-x}{e^{-x}-1} + \frac{-x}{2}\right) \\ &= \frac{x}{e^x-1} + \frac{x}{2} + \frac{x}{e^{-x}-1} + \frac{x}{2}\\ &= \frac{x}{e^x-1} + \frac{e^x x}{1-e^{x}} + x = \frac{x}{e^x-1} - \frac{e^x x}{e^x-1} + x\\ &= \frac{x - e^x x}{e^x-1} + x = \frac{x(1- e^x)}{e^x-1} + x = -x + x \\&= 0 \end{align}$$ showing that $f(x)=f(-x)$.

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Quick note: the expression of $f$ gives that $f$ is defined on $\mathbb{R}^\ast$, although it can be extended by continuity at $0$ (with $f(0)=1$).

Answer 2

Hint : $\frac{1}{e^{-x}-1}=\frac{e^x}{1-e^x}$