For this integral: $\int_0^{\frac{\pi}{4}} x \tan(x) dx$

Question

For the following integral: $$\int_0^{\frac{\pi}{4}} x \tan(x) dx=\frac{G}{2}-\frac{\pi\ln(2)}{8}$$ where $G$ is Catalan's constant. After integrating by parts, it is equivalent to compute $\int_0^{\pi/4} \ln(\cos(x)) dx$. How can I proceed further?

Answer 1

We want to evaluate the definite integral $$ \mathcal{I} = \displaystyle \int_0^{\frac{\pi}{4}} \ln( \cos x ) \ \mathrm dx $$

We note that $$ \begin{align} \ln(\cos{x}) &=\frac12\Big( ( \ln(\sin{x}) + \ln(\cos{x}) ) -(\ln(\sin{x}) -\ln(\cos{x}))\Big) \\ &= \frac12 \ln \Big( \frac{\sin(2x)}{2}\Big) -\frac12\ln(\tan{x}) \\ &= \frac12\ln(\sin(2x)) -\frac12 \ln(2) - \frac12\ln(\tan{x}) \end{align}$$

Using this, the integrand simplifies to

$$ \mathcal{I} =\frac12 \displaystyle \int_0^{\frac{\pi}{4}} \ln(\sin(2x))\ \mathrm dx - \frac12 \int_0^{\frac{\pi}{4}} \ln(\tan{x})\ \mathrm dx -\frac12\int_0^{\frac{\pi}{4}} \ln{2} \ \mathrm dx $$

Starting with the second integral, it is a well known result.

$$ \int_0^{\frac{\pi}{4}} \ln(\tan{x})\ \mathrm dx = -G$$

$ G$ denotes Catalan's constant.

For the first integral, the substitution $ 2x=t $ gives $$\dfrac{1}{2}\int_0^{\frac{\pi}{2}} \ln(\sin{x})\ \mathrm dx $$ This integral just equals $ - \dfrac{\pi}{4}\ln{2}$, by using the well known result

$$ \int_0^{\frac{\pi}{2}} \ln(\sin{x})\ \mathrm dx = - \dfrac{\pi}{2}\ln{2}$$

The third integral is trivial and equals $ \frac{\pi}{4}\ln{2}$.

Summing up the values of the 3 integrals, our original integral equals

$$ \boxed{\boxed{\int_0^{\frac\pi4}\ln(\cos x )\,\mathrm dx =\frac G2 - \dfrac{\pi}{4}\ln{2}}} $$

And using integration by parts,

$$\begin{align}\int_0^{\frac\pi4}x\tan x\,\mathrm dx &= \frac\pi8\ln2+\int_0^{\frac\pi4}\ln(\cos x)\,\mathrm dx \\ &= \frac G2-\frac\pi8\ln2\end{align}$$ Which matches the result given in the OP.

EDIT

As people had problems with the second integral, here is solution.

The Catalan's constant is defined as

$$G=\beta(2)= \sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}$$

Here $\beta(\cdot)$ denotes the Dirichlet's beta function.

In our integral, we substitute $\tan x = t$.

$$\begin{align}\int_0^{\frac\pi4}\ln \tan x\,\mathrm dx &= \int_0^1 \frac{\ln t}{1+t^2}\,\mathrm dt \\ &= \sum_{k=0}^\infty(-1)^k \int_0^1 x^{2k}\ln x\,\mathrm dx \\ &= -\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2}\\ &= -G\end{align}$$

Answer 2

Using the complex definition of cosine,

$$\begin{align}\ln\cos x&= \ln\Big(\frac{e^{ix}+e^{-ix}}2\Big) \\ &= ix+\ln(1+e^{-2ix})-\ln2 \\ &= -ix+\ln(1+e^{2ix})-\ln2 \\ \implies \ln\cos x &= -\ln 2 +\frac12(\ln(1+e^{2ix})+\ln(1+e^{-2ix})) \\ &= -\ln 2 -\frac12 \sum_{k=1}^\infty (-1)^k\frac1k(e^{2ikx}+e^{-2ikx}) \\ \ln\cos x&= -\ln2-\sum_{k=1}^\infty \frac{(-1)^k}k\cos(2kx) \end{align}$$

This is the Fourier series of $\ln\cos x$. Using this result,

$$\begin{align}I =\int_0^{\pi/4}\ln\cos x\,\mathrm dx &= \int_0^{\pi/4}-\ln2 -\sum_{k=1}^\infty\frac{(-1)^k}k \cos(2k x)\,\mathrm dx \\ &= -\frac\pi4\ln2-\sum_{k=1} ^\infty \frac{(-1)^k}k\int_0^{\pi/4}\cos(2kx)\,\mathrm dx \\ &= -\frac\pi4\ln2-\frac12\sum_{k=1}^\infty \frac{(-1)^k}{k^2} \sin\Big(\frac{k\pi}2\Big) \end{align}$$

Now, we note that for $k\in\mathbb Z$

$$\sin(k\pi) =0 , \quad \sin\Big(\frac{(2k+1)\pi}2\Big) = (-1)^k $$

Using this, we re-index our sum.

$$\begin{align}I &= -\frac\pi4\ln2-\frac12\sum_{k=0}^\infty\frac{(-1)^{2k+1}(-1)^k}{(2k+1)^2} \\ &= -\frac\pi4\ln2+\frac12\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2} \\ I &= \frac G2-\frac\pi4\ln2 \end{align}$$

Answer 3

To the nice solutions - just to add another approach (which does not use the trigonometry): $$I=\int_0^{\pi/4}x\tan xdx=(x=\tan^{-1}t)\,\,\int_0^1\tan^{-1}t\frac{tdt}{1+t^2}$$ $$= (IBP)\,\,\,\,\frac{\pi\ln2}{8}\ln2-\frac{1}{2}\int_0^1\frac{\ln(1+t^2)}{1+t^2}dt$$ Making change $x=1/t$ $$\int_0^1\frac{\ln(1+t^2)}{1+t^2}dt=\int_1^\infty\frac{\ln(1+x^2)}{1+x^2}dx-2\int_1^\infty\frac{\ln x}{1+x^2}dx$$ $$2\int_0^1\frac{\ln(1+t^2)}{1+t^2}dt=\int_0^\infty\frac{\ln(1+x^2)}{1+x^2}dx-2G$$ The last integral it is easy to evaluate via complex integration: $$\int_0^\infty\frac{\ln(1+x^2)}{1+x^2}dx=\Re\int_{-\infty}^\infty\frac{\ln(1-ix)}{1+x^2}dx$$ Closing the contour in the upper half-plane (where we have one simple pole and no branch points for the chosen integrand) $$\int_{-\infty}^\infty\frac{\ln(1-ix)}{1+x^2}dx=2\pi i\operatorname{Res}_{x=i}\frac{\ln(1-ix)}{1+x^2}=2\pi i\frac{\ln2}{2i}=\pi\ln2$$ Taking all together $$I=\frac{\pi\ln2}{8}-\frac{\pi\ln2}{4}+\frac{G}{2}=\frac{G}{2}-\frac{\pi\ln2}{8}$$

Answer 4

By integration by parts, we have

$$ \begin{aligned} \int_{0}^{\frac{\pi}{4}} x \tan x d x &=-\int_{0}^{\frac{\pi}{4}} x d(\ln (\cos x)) \\ &=-\left[ x \ln (\cos x)\right]_{0}^{\frac{\pi}{4}}+\int_{0}^{\frac{\pi}{4}} \ln (\cos x) d x\\&= -\frac{\pi}{4} \ln \left(\frac{1}{\sqrt{2}}\right)+\int_{0}^{\frac{\pi}{4}} \ln (\cos x) d x \end{aligned} $$

By my post in Quora,

$$\int_{0}^{\frac{\pi}{4}} \ln (\cos x) dx =-\frac{\pi}{4} \ln 2+\frac{G}{2},\tag*{} $$ $\textrm{where G is the Catalan's constant.}$

Hence we can conclude that

$$\begin{aligned}\int_{0}^{\frac{\pi}{4}} x \tan x d x &= -\frac{\pi}{4} \ln \left(\frac{1}{\sqrt{2}}\right) -\frac{\pi}{4} \ln 2+\frac{G}{2}\\&= -\frac{\pi}{8} \ln 2+\frac{G}{2}\end{aligned}$$

Answer 5

Rewrite the integral as \begin{align} \int_{0}^{\frac{\pi}{4}} x \tan x d x &=-\int_{0}^{\frac{\pi}{4}} x d[\ln (2\cos x)] \overset{ibp}= -\frac{\pi}{8} \ln2 +I \end{align} where $I=\int_{0}^{\frac{\pi}{4}} \ln (2\cos x) dx $, along with $J=\int_{0}^{\frac{\pi}{4}} \ln (2\sin x) dx $. Note \begin{align} I-J&=-\int_0^{\frac\pi4}\ln (\tan x )dx=G\\ I+J &= \int_{0}^{\frac{\pi}{4}} \ln (2\sin 2x) dx \overset{x\to\frac\pi4 -x}= \int_{0}^{\frac{\pi}{4}} \ln (2\cos 2x) dx\\ &=\frac12 \int_{0}^{\frac{\pi}{4}} \ln (4\sin 2x\cos2x) dx \overset{2x\to x}= \frac14 \int_{0}^{\frac{\pi}{2}} \ln (2\sin 2x) dx\\ &= \frac12 (I+J)=0 \end{align} which leads to $I=\frac12G$ and $$\int_{0}^{\frac{\pi}{4}} x \tan x d x = \frac12G -\frac{\pi}{8} \ln2 $$

Answer 6

Another way, \begin{align}J&=\int_0^{\frac{\pi}{4}}t\tan t dt\\ &\overset{x=\tan t}=\int_0^1 \frac{x\arctan x}{1+x^2}dx\\ &=\int_0^1 \frac{x}{1+x^2}\left(\underbrace{\int_0^x \frac{1}{1+u^2}du}_{z=\frac{u}{x}}\right)dx\\ &=\int_0^1 \int_0^1 \frac{x^2}{(1+x^2)(1+x^2z^2)}dxdz\\ &=\int_0^1 \int_0^1 \left(\frac{1}{(1-z^2)(1+x^2z^2)}-\frac{1}{(1-z^2)(1+x^2)}\right)dxdz\\ &=\int_0^1 \int_0^1 \frac{1}{z(1-z^2)}\left(\frac{z}{1+x^2z^2}-\frac{z}{1+x^2}\right)dxdz\\ &=\int_0^1 \frac{1}{z(1-z^2)}\left(\left(\underbrace{\int_0^1 \frac{z}{1+x^2z^2}dx}_{u=xz}\right)-z\int_0^1\frac{1}{1+x^2}dx\right)dz\\ &=\int_0^1 \frac{1}{z(1-z^2)}\left(\left(\int_0^z \frac{1}{1+u^2}du\right)-\frac{\pi z}{4}\right)dz\\ &\overset{\text{IBP}}=\underbrace{\left[\left(\ln z-\frac{1}{2}\ln(1-z^2)\right)\left(\left(\int_0^z \frac{1}{1+u^2}du\right)-\frac{\pi z}{4}\right)\right]_0^1}_{=0}-\\&\int_0^1 \left(\ln z-\frac{1}{2}\ln(1-z^2)\right)\left(\frac{1}{1+z^2}-\frac{\pi}{4}\right)dz\\ &=-\int_0^1 \frac{\ln z}{1+z^2}dz+\frac{\pi}{4}\int_0^1 \ln z dz+\frac{1}{2}\int_0^1 \frac{\ln(1-z^2)}{1+z^2}dz-\frac{\pi}{8}\int_0^1 \ln(1-z^2)dz\\ &=\text{G}-\frac{\pi}{4}+\frac{1}{2}\int_0^1 \frac{\ln(1-z^2)}{1+z^2}dz-\frac{\pi}{8}\underbrace{\left[x\ln(1-x^2)-\ln\left(\frac{1-x}{1+x}\right)-2x\right]_0^1}_{=2\ln 2-2}\\ &=\text{G}+\frac{1}{2}\underbrace{\int_0^1 \frac{\ln(1-z^2)}{1+z^2}dz}_{y=\frac{1-z}{1+z}}-\frac{1}{4}\pi\ln 2\\ &=\text{G}+\frac{1}{2}\int_0^1 \frac{\ln\left(\frac{4y}{(1+y)^2}\right)}{1+y^2}dy-\frac{1}{4}\pi\ln 2\\ &=\text{G}+\int_0^1 \frac{\ln 2}{1+y^2}dy+\frac{1}{2}\int_0^1 \frac{\ln y}{1+y^2}dy-\int_0^1 \frac{\ln(1+y)}{1+y^2}dy-\frac{1}{4}\pi\ln 2\\ &=\frac{1}{2}\text{G}-\underbrace{\int_0^1 \frac{\ln(1+y)}{1+y^2}dy}_{=\text{K}}\\ \text{K}&\overset{z=\frac{1-y}{1+y}}=\int_0^1 \frac{\ln\left(\frac{2}{1+z}\right)}{1+z^2}dz=\frac{1}{4}\pi\ln 2-\text{K}\\ \text{K}&=\frac{1}{8}\pi\ln 2\\ J&=\boxed{\frac{1}{2}\text{G}-\frac{1}{8}\pi\ln 2} \end{align}