Evaluating the Integral Using the Clausen Function

[DreamWeaver]

Find a closed form evaluation for the following trigonometric integral, where the \(\displaystyle 0 < \theta \le \pi/2\):\(\displaystyle \int_0^{\theta}\frac{x^2}{\sin x} \, dx= \text{?}\)

Hint:

Spoiler

Consider

\(\displaystyle \int_0^{\theta} x\log \left(\tan \frac{x}{2} \right)\, dx\)

and then express this logtangent integral in terms of Clausen functions, by splitting logtan into logsin + logcos integrals...

 

[Shobhit]

Hi all, this is my first post on Math Help Boards. :)

The answer to this problem is
$$\theta^2 \log \tan \frac{\theta}{2} -\frac{7}{2}\zeta(3)+4\theta \text{Cl}_2(\theta)-\theta \text{Cl}_2(2\theta)+4\text{Cl}_3(\theta)-\frac{1}{2}\text{Cl}_3(2\theta)$$

where $\text{Cl}_n(z)$ denotes the Clausen Function.

Proof:

Spoiler

Let $I$ denote our integral. On applying integration by parts we obtain

$$
I=\theta^2 \log \tan \frac{\theta}{2}-2\int_0^{\theta}x \log \tan \frac{x}{2}\; dx
$$

Now, we invoke the Fourier series of $\log \tan \frac{x}{2}$:

$$\log \tan \frac{x}{2}=-2 \sum_{n=1}^\infty \frac{\cos(2n-1)x}{2n-1}$$

It follows that

$$
\begin{align*}
I &= \theta^2 \log \tan \frac{\theta}{2}+4\sum_{n=1}^\infty \frac{1}{2n-1}\int_0^{\theta}x \cos(2n-1)x \; dx \\
&= \theta^2 \log \tan \frac{\theta}{2}+4\sum_{n=1}^\infty \frac{1}{2n-1}\left\{\theta \frac{\sin(2n-1)\theta}{2n-1}-\frac{1}{2n-1}\int_0^\theta \sin(2n-1)x dx\right\} \\
&= \theta^2 \log \tan \frac{\theta}{2}+4\sum_{n=1}^\infty \frac{1}{2n-1}\left\{\theta \frac{\sin(2n-1)\theta}{2n-1}+\frac{\cos(2n-1)\theta -1}{(2n-1)^2}\right\} \\
&= \theta^2 \log \tan \frac{\theta}{2} -\frac{7}{2}\zeta(3) +4\theta \sum_{n=1}^\infty \frac{\sin(2n-1)\theta}{(2n-1)^2}+4\sum_{n=1}^\infty \frac{\cos (2n-1)\theta}{(2n-1)^3} \\
&= \theta^2 \log \tan \frac{\theta}{2} -\frac{7}{2}\zeta(3)+4\theta \text{Cl}_2(\theta)-\theta \text{Cl}_2(2\theta)+4\text{Cl}_3(\theta)-\frac{1}{2}\text{Cl}_3(2\theta)
\end{align*}
$$

In the last step, I used

$$
\begin{align*}\sum_{n=1}^\infty \frac{\cos(2n-1)\theta}{(2n-1)^3} &=\text{Cl}_3(\theta)-\frac{1}{8}\text{Cl}_3(2\theta) \\
\sum_{n=1}^\infty \frac{\sin(2n-1)\theta}{(2n-1)^2}&=\text{Cl}_2(\theta)-\frac{1}{4}\text{Cl}_2(2\theta)
\end{align*}
$$