How can I evaluate this improper integral?

Question

I have a problem: evaluate $$\int_{0}^{\infty}\frac{\cos(x)-\cos(2x)}{x}dx\,.$$ I am told this integral is not an elementary one and that's why I am stuck where to start. Thank you for helping me.

Answer 1

$$\int_a^{\infty}\frac{\cos(2x)}{x}dx = \int_{2a}^{\infty}\frac{\cos(x)}{x}dx$$ and so this integral equals $$\lim_{a\downarrow 0}\int_a^{2a}\frac{\cos(x)}{x}dx$$ which is easily found to be equal to $\log(2)$ using the inequalities $$1-\frac{x^2}{2}\leq\cos(x)\leq 1.$$

Answer 2

$$ \int_{0}^{\infty}\frac{\cos x-\cos 2x}{x}=\int_{0}^{\infty}(\mathscr L(\cos x)-\mathscr L(\cos 2x)) \\ =\int_{0}^{\infty}(\frac{s}{s^{2}+1}-\frac{s}{s^{2}+4})ds=\frac 12 \ln \left( \frac{s^{2}+1}{s^{2}+4}\right)_{0}^{\infty} =\ln 2 $$

Answer 3

This integral can be written $$ \mathrm{Re}\left(\int_{0}^{\infty}\frac{e^{ix}-e^{2ix}}{x}\,dx\right) $$ We move the contour of integration so that we integrate along the positive imaginary axis instead of the positive real axis (it can be checked that the integrays decays quickly enough at infinity for this to be valid). Rewriting our new integral with $x=it$ gives $$ \int_0^\infty t^{-1}(e^{-t}-e^{-2t})\,dt $$ If we replace the $t^{-1}$ with $t^{-1+\epsilon}$, for some $\epsilon>0$, the value of the above integral is $\Gamma(\epsilon)(1-2^{-\epsilon})$. As $\epsilon\to 0$, $\Gamma(\epsilon)\sim \frac{1}{\epsilon}$, so the limit of the above expression is $\log(2)$

Answer 4

This is a bit similar to @aliakbar's way but in detailed. Let $f(x)=\cos(x)-\cos(2x)$ and let's evaluate the following integral: $$I(s)=\int_{0}^{\infty}\exp(-sx)\frac{f(x)}{x}~dx$$ We can easily seen that $\lim_{x\to 0^+}\frac{f(x)}{x}=0$ and that: $$\mathcal{L}\{\cos(x)-\cos(2x)\}=\frac{s}{s^2-1}-\frac{s}{s^2+4}$$ Moreover $$\int_0^{\infty}\frac{f(x)}{x}=\int_0^{\infty}F(s)~ds$$ wherein $F(s)=\mathcal{L(f(x))}$. So $I(s)=\mathcal{L\left(\frac{f(x)}{x}\right)}=\int_s^{\infty}\left(\frac{t}{t^2-1}-\frac{t}{t^2+4}\right)dt=...=\frac{1}{2}\ln\left(\frac{s^2+4}{s^2+1}\right)$. Now set $s=0$ in the later integral. It equals to $\ln(2)$.

Answer 5

You have to use the limited developement of cos(x) and cos(2x).