Why is $\int_0^{2\pi}\frac{x\cos x}{1+\cos x}dx$ not zero?

Question

I want to evaluate the following integral:

$$\int_0^{2\pi}\frac{x\cos x}{1+\cos x}dx$$

In this, first I replace $x$ with $2\pi -x$. $$ I = \pi\int_{0}^{2\pi}{\frac{\cos(x)}{1+\cos(x)}\,\mathrm{d}x}$$ After that I thought that both limits $0$ and $2\pi$ are the same angle for cosine so integration should be zero.

Answer 1

So you are asked to evaluate the integral $\int_0^{2\pi}\frac{x\cos x}{1+\cos x}dx$ and this question comes from a textbook ...

IMHO, there are two options :

• this is merely an error, and this happens sometimes, even in good textbooks

• the author of the questions wants you to answer that it is not possible to evaluate the given integral, since it is a divergent one !

Here is why ...

The denominator $1+\cos(x)$ vanishes at $x=\pi$ and so, the convergence of this integral would mean the convergence of :

$$\int_0^{\pi}\frac{x\cos x}{1+\cos x}dx\quad\color{red}{\textrm{ and }}\quad\int_\pi^{2\pi}\frac{x\cos x}{1+\cos x}dx$$

But both diverge, because :

$$\frac{x\cos(x)}{1+\cos(x)}\underset{x\to\pi}{\sim}\frac{-\pi}{1-\cos(x-\pi)}\underset{x\to\pi}{\sim}\frac{-2\pi}{(x-\pi)^2}$$

[ The last equivalence comes from $\cos(h)\underset{h\to0}{=}1-\frac{h^2}{2}+o(h^2)$ ]

and we know that $\displaystyle{\int_0^1\frac{dt}{t^\alpha}}$ converges iff $\alpha<1$.

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That said, it would be interesting to compute :

$$A=\int_0^{\pi/2}\frac{x\cos(x)}{1+\cos(x)}\,dx$$

which is well defined !

And the answer is $\displaystyle{A=\frac{\pi^2}8-\frac\pi2+\ln(2)}$.

Answer 2

No, since $x$ is also replaced by $2x-\pi$. So you get $$I = \int_{0}^{2\pi}{\frac{2\pi\cos(x)}{1+\cos(x)}\,\mathrm{d}x}-I$$

And $$I = \pi\int_{0}^{2\pi}{\frac{\cos(x)}{1+\cos(x)}\,\mathrm{d}x}$$