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

definite integralsintegration

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?

Best Answer

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 $\tan x = t$.

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

Related Question