3
$\begingroup$

How to prove $\int^{\pi}_0 \cos(x)\log(\tan(\frac{x}{4}))\,\mathrm dx = -2$?

Or $\int^1_0 \log(u)\frac{(u^4-6u^2+1)}{(u^2+1)^3} \,\mathrm du = -1/2$.

(I substituted $\tan(x/4) = u$.)

$\endgroup$
5
$\begingroup$

\begin{align*} \int_0^\pi\cos x\log\left(\tan \frac x4\right)dx&=\left[\sin x\log\left(\tan\frac x4\right)\right]_0^\pi-\int_0^\pi\sin x\cdot\frac{\frac14\sec^2\frac x4}{\tan\frac x4}dx\\ &=0-\int_0^\pi\frac{\sin x}{4\sin\frac x4\cos\frac x4}dx\\ &=-\int_0^\pi\frac{\sin x}{2\sin\frac x2}dx\\ &=-\int_0^\pi\cos\frac x2dx\\ &=-\left[2\sin\frac x2\right]_0^\pi\\ &=-2 \end{align*}

Note that

\begin{align*} \lim_{x\to0^+}\sin x\log\left(\tan\frac x4\right)&=\lim_{x\to0^+}\frac{\log\left(\tan\frac x4\right)}{\csc x}\\ &=\lim_{x\to0^+}\frac{\frac{\sec^2\frac x4}{4\tan\frac x4}}{-\cot x\csc x}\\ &=\lim_{x\to0^+}\left(-\sin x\tan x\cos\frac x2\right)\\ &=0 \end{align*}

$\endgroup$
  • $\begingroup$ @PeterForeman Yes, this is necessary. $\endgroup$ – CY Aries Jul 7 at 10:06
2
$\begingroup$

$\int_{0}^{\pi} \cos{x}\log(\tan(x/4)) dx $

$= \int_{0}^{\pi/4} \cos{4z}\log(\tan(z)) dz $ (where $z = x/4$)

$= [\sin{4z}\log(\tan(z))]_0^{\pi/4} - \int_{0}^{\pi/4} \frac{\sin{4z}\sec^2{z}}{\tan{z}} dz $

$= -\int_{0}^{\pi/4} \frac{\sin{4z}}{\sin{z}\cos{z}} dz$

$=-4\int_{0}^{\pi/4} \cos{2z}dz = -2$

$\endgroup$
2
$\begingroup$

The integral you got after the substitution can be solved using integration by parts. That is, $$\int_0^1 f(u) g'(u)du=f(u)g(u)|_0^1 - \int_0^1 f'(u) g(u) du,$$ where $$f(u)=\log(u)$$ and $$g'(u)=\frac{u^4-6u^2+1}{(u^2+1)^3}.$$ In this case, $f'(u)=\frac1u$ and $g(u)$ can be obtained integrating $g'(u)$ using simple fractions. I'm not saying it will be fast nor nice to do it; just that it can be done.

NOTE: I haven't tried it, but it could be that after the substitution $u=\frac x4$, and after replacing $$\cos(4u)=\cos^2(2u)-\sin^2(2u)=$$$$=(\cos^2(u)-\sin^2(u))^2-(2\sin(u)\cos(u))^2=$$$$=\cos^4(u)-6\sin(u)\cos(u)+\sin^4(u),$$ and separating into the sum of three integrals, each one of them can be solved integrating by parts, where the $\log$ part will be $f$, that is, the function you'll take the derivative to. But again, I don't really know if this will work.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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