Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Here is a related problem. –  Mhenni Benghorbal Jan 3 '13 at 1:44
add comment

5 Answers

up vote 12 down vote accepted

$$\int_a^{\infty}\frac{\cos(2x)}{x}dx = \int_{2a}^{\infty}\frac{\cos(x)}{x}dx$$ 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 $$1-\frac{x^2}{2}\leq\cos(x)\leq 1.$$

share|improve this answer
1  
+1: I believe this answer gives the best intuition what is going on here... –  Fabian Jan 1 '13 at 18:52
    
In principle there might be an analogous contribution to the result, equal to the limit when $b\to\infty$ of the integral of $x\mapsto(\cos x)/x$ on $(b,2b)$. One should show that actually this contribution disappears. –  Did Jan 1 '13 at 19:18
    
@did Yes, absolutely. That will follow from the fact that $\int_a^{\infty}\frac{\cos(x)}{x}dx$ is a well defined improper integral to begin with. –  WimC Jan 1 '13 at 19:29
add comment

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

share|improve this answer
2  
The integrals on the third line do not converge. –  Fabian Jan 1 '13 at 18:28
    
The answer is edited. –  aliakbar Jan 1 '13 at 18:33
add comment

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 $\log(2)$

share|improve this answer
    
The final result $\log2$ is definitely correct but the juggling with real/imaginary exponentials to get it is not quite convincing at the moment. –  Did Jan 1 '13 at 18:20
add comment

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 $\ln(2)$.

share|improve this answer
add comment

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

share|improve this answer
    
You mean the series expansion? –  Pedro Tamaroff Jan 1 '13 at 18:27
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.