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By some analysis and through Wolfram|Alpha I know that the integral in question is equal to a fascinating $$I=\int\limits_{0}^{\frac{\pi}{4}}x\ln\left(1+\tan x\right)dx=\frac{21}{64}\zeta(3)+\frac{\pi^{2}}{64}\ln 2-\frac{\pi}{8}G$$ where $G$ is Catalan's constant.

However, I have tried many methods to evaluate this integral, all to no avail. Using the Maclaurin series for $\ln(1+\tan x)$ unfortunately produces $$I=\sum\limits_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{k+1}\int\limits_{0}^{\frac{\pi}{4}}x\tan^{k+1}xdx$$ the integral in which is particularly hard to deal with - spawning this question of mine, the answers and comments to which destroyed my dreams of continuing down this path.

Alternatively, differentiating under the integral sign with $$I(n)=\int\limits_{0}^{\frac{\pi}{4}}x\ln\Big(\tan\left(nx\right)+1\Big)dx\Rightarrow I'(n)=\int\limits_{0}^{\frac{\pi}{4}}\frac{x}{n+\tan x}dx$$ looks equally hopeless after a few calculations.

Integration by parts does not seem to work very well because the antiderivative of $\ln\left(1+\tan x\right)$ is absolutely hideous.

Perhaps Clausen functions can help? Or maybe the Fourier series of $\ln\left(1+\tan x\right)$ which I unfortunately am unfamiliar with...

In any case, the existence of $\zeta(3)$ and $G$ in the answer scream an infinite sum, and the fractional coefficients also hint at some substitutions that could aid us along the way - but my ideas stop here.

Any insights are greatly appreciated.

Kisaragi Ayami
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3 Answers3

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The trick here seems to be the well-known Fourier series $$ \log(2\cos x) = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{\cos(2kx)}{k}, \hspace{0.5cm} -\frac{\pi}{2}<x<\frac{\pi}{2} $$ This can be found by letting $z=re^{ix}$ into the Taylor series for the logarithm, $$ \log(1+z) = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{z^k}{k}, \hspace{0.5cm} |z|<1 $$ and then letting $r \to 1^{-}$. The resuling series converges for $-\frac{\pi}{2}<x<\frac{\pi}{2} $ by Dirichlet's test. Replacing $x$ with $x-\frac{\pi}{4}$ gives the Fourier series $$ \log(2\cos(x-\tfrac{\pi}{4})) = \log(\sqrt{2} (\cos x+\sin x)) = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{\cos(2kx+\frac{\pi k}{2})}{k} $$ $$ = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{\cos(4kx)}{2k} + \sum_{k=0}^{\infty} (-1)^k \frac{\sin(2(2k+1)x)}{2k+1}, \hspace{0.5cm} -\frac{\pi}{4}<x<\frac{3\pi}{4} $$ Subtracting the two series gives the desired expression for $\log(1+\tan x)$: $$ -\frac{1}{2} \log 2 + \log(1+\tan x) = \log \left(\frac{\sqrt{2}}{2} \frac{\cos x+\sin x}{\sin x} \right) $$ $$ = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{\cos(4kx)}{2k} + \sum_{k=0}^{\infty} (-1)^k \frac{\sin(2(2k+1)x)}{2k+1} - \sum_{k=1}^{\infty} (-1)^{k+1} \frac{\cos(2kx)}{k}, \hspace{0.5cm} 0<x<\frac{\pi}{2} $$ Now multiply through by $x$ and integrate term-by-term. Can you finish it from here?

Dave
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7

If we write $$\log (1+\tan (x))=\log (\sin (x)+\cos (x))-\log (\cos (x))$$ and use Euler representation of the sine and cosine, the antiderivative write $$\int x\log (1+\tan (x))\,dx=\frac{1}{2} x^2 \log (1+i)-$$ $$\frac{1}{2} i x \left(\text{Li}_2\left(-e^{2 i x}\right)-\text{Li}_2\left(i e^{2 i x}\right)\right)+\frac{1}{4} \left(\text{Li}_3\left(-e^{2 i x}\right)-\text{Li}_3\left(i e^{2 i x}\right)\right)$$ $$I=\int_0^{\frac \pi 4} x\log (1+\tan (x))\,dx=$$ $$\frac{1}{64} \left(-8 \pi C+16 \text{Li}_3(-i)+12 \zeta (3)+\pi ^2 \log (2)\right)-\left(-\frac{3 \zeta (3)}{16}-\frac{\text{Li}_3(i)}{4} \right)$$ $$I=\frac{21}{64}\zeta(3)+\frac{\pi^{2}}{64}\log( 2)-\frac{\pi}{8}C$$

Claude Leibovici
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\begin{align} I=&\int_{0}^{\frac{\pi}{4}}x\ln\left(1+\tan x\right)dx\\ =& \int_{0}^{\frac{\pi}{4}}x\ln\frac{4(1+\tan x)}{\sec^2 x}dx -2\int_{0}^{\frac{\pi}{4}}x\ln\left(2\cos x\right)dx \end{align} where $ \int_{0}^{\frac{\pi}{4}}x\ln\left(2\cos x\right)dx = \frac{\pi}{8}G-\frac{21}{128} \zeta(3)$ and \begin{align} J= &\int_{0}^{\frac{\pi}{4}}x\ln\frac{4(1+\tan x)}{\sec^2 x}\overset{x\to\frac\pi4-x}{dx} =\frac\pi4 \int_{0}^{\frac{\pi}{4}}\ln\frac{4(1+\tan x)}{\sec^2 x}dx-J\\ =& \ \frac\pi8 \bigg(\int_{0}^{\frac{\pi}{4}}\ln(1+\tan x)dx +\int_{0}^{\frac{\pi}{4}}\ln(4\cos^2 x)dx\bigg)\\ =&\ \frac\pi8\bigg( \frac\pi8\ln2+G \bigg)= \frac{\pi^2}{64}\ln2+\frac\pi8G\\ \end{align} Substitute above results into $I$ to obtain $$I=\frac{\pi^{2}}{64}\ln2-\frac{\pi}{8}G+ \frac{21}{64}\zeta(3) $$

Quanto
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