[TUT] Integrals involving log. and trig. functions

[Tolaso J Kos]

So that we combine this post with some other interesting integrals.

1. $\displaystyle \int_{0}^{\pi/4} \log^n \left ( \tan x \right )\, {\rm d}x = (-1)^n n! \beta(n+1)$

Proof: We have derived it to the linked post - first integral. Here $\beta$ denotes the Beta Dirichlet function.

2. $\displaystyle \int_{0}^{\pi/2} \log^n \left ( \tan x \right )\, {\rm d}x= \left\{\begin{matrix}
0& , & n \; \text{odd}\\
2n! \beta(n+1)&, &n \; \text{even}
\end{matrix}\right.$

Proof: Make the change of variables $u=\tan x$ and you get exactly the first integral in the linked post.

3. $\displaystyle \int_{0}^{\theta} \log^n \left ( \tan x \right )\, {\rm d}x = n! \sum_{k=0}^{n}(-1)^k \frac{\log^{n-k} \left ( \tan \theta \right )}{\left ( n-k \right )!} {\rm Ti}_{k+1} \left ( \tan \theta \right )$

Proof: The proof is omitted. Anyone is welcome to try it. Here ${\rm Ti}_k$ denote the inverse tangent integral, defined recursively as:

$${\rm Ti}_1 (z)= \frac{\arctan z}{z} \quad , \quad {\rm Ti}_{m+1}(z)= \int_{0}^{z} \frac{{\rm Ti}_m(z)}{z}\, {\rm d}z$$

4. $\displaystyle \int_{0}^{\theta} \log \left ( \sin x \right )\, {\rm d}x = -\frac{1}{2}{\rm Cl}_2 (2\theta) - \theta \log 2$ and $\displaystyle \int_{0}^{\theta} \log \left ( \cos x \right )\, {\rm d}x = \frac{1}{2}{\rm Cl}_2 (\pi -2\theta) - \theta \log 2$.

Proof: Invoke Fourier series of $\log \sin x$ and $\log \cos x$ respectively. Here ${\rm Cl}_2$ denotes the Clausen functions.

Feel free to complete this short tutorial or give some proofs for the results presented here.

[Tolaso J Kos]

Since no - one wanted to chew on any of the proposed questions I'll give them a shot.

$$ \int_{0}^{\theta} \log \left ( \sin x \right )\, {\rm d}x = -\frac{1}{2}{\rm Cl}_2 (2\theta) - \theta \log 2$$

The Fourier series of $\log \sin x$ is

$$\log \sin x = - \sum_{k=1}^{\infty} \frac{\cos 2kx}{k} - \log 2$$

thus:

\begin{align*}
\int_{0}^{\theta} \log \sin x \, {\rm d}x &= \int_{0}^{\theta} \left ( -\log 2 - \sum_{k=1}^{\infty} \frac{\cos 2kx}{k} \right )\, {\rm d}\theta \\
&= - \theta \log 2 - \int_{0}^{\theta} \sum_{k=1}^{\infty} \frac{\cos 2kx}{k}\, {\rm d}x\\
&= -\theta \log 2 - \sum_{k=1}^{\infty} \frac{1}{k} \int_{0}^{\theta} \cos 2kx \, {\rm d}x\\
&= - \theta \log 2 - \frac{1}{2}\sum_{k=1}^{\infty} \frac{\sin 2 \theta k}{k^2} \\
&=-\theta \log 2 - \frac{1}{2} {\rm Cl}_2 \left ( 2\theta \right )
\end{align*}

Similarly using the Fourier series of $\log \cos x$ we get the other formula. The Fourier series of $\log \cos x$ is

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

$$\int_{0}^{\theta} \log \left ( \cos x \right )\, {\rm d}x = \frac{1}{2}{\rm Cl}_2 (\pi -2\theta) - \theta \log 2$$

Basic stuff about the Clausen Functions may be found here.

[Tolaso J Kos]

$$ \int_{0}^{\theta} \log^n \left ( \tan x \right )\, {\rm d}x = n! \sum_{k=0}^{n}(-1)^k \frac{\log^{n-k} \left ( \tan \theta \right )}{\left ( n-k \right )!} {\rm Ti}_{k+1} \left ( \tan \theta \right )$$

We have successively:

\begin{align*}
\int_{0}^{\theta} \log^m \tan x \, {\rm d}x &=\int_{0}^{\tan \theta} \frac{\log^m x}{x^2+1}\, {\rm d}x \\
&= \int_{0}^{\tan \theta} \log^m x \sum_{k=0}^{\infty} (-1)^k x^{2k}\, {\rm d}x \\
&=\sum_{k=0}^{\infty} (-1)^k \int_{0}^{\tan \theta} x^{2k} \log^m x \, {\rm d}x \\
&\!\!\!\!\!\!\!\!\!\!\!\overset{x=\tan \theta y}{=\! =\! =\! =\! =\! =\! =\!}\sum_{k=0}^{\infty}(-1)^k \tan^{2k+1} \theta \int_{0}^{1} y^{2k} \left (\log \tan \theta + \log y \right )^m \, {\rm d}y \\
&= \sum_{j=0}^{m} \binom{m}{j} \left ( \log \tan \theta \right )^{m-j} \sum_{k=0}^{\infty} (-1)^k \tan^{2k+1} \theta \int_{0}^{1}y^{2k} \log^j y \, {\rm d}y \\
&=\sum_{j=0}^{m} (-1)^j j! \binom{m}{j} \log^{m-j} \tan \theta \sum_{k=0}^{\infty} \frac{(-1)^k (\tan \theta)^{2k+1}}{\left ( 2k+1 \right )^{j+1}} \\
&=m! \sum_{j=0}^{m} (-1)^j \frac{\left ( \log \tan \theta \right )^{m-j}}{\left ( m-j \right )!} \sum_{k=0}^{\infty} \frac{(-1)^k \left ( \tan \theta \right )^{2k+1}}{\left ( 2k+1 \right )^{j+1}} \\
&=m! \sum_{j=0}^{m} (-1)^j \frac{\log^{m-j} \tan \theta}{(m-j)!} {\rm Ti}_{j+1} (\tan \theta)
\end{align*}

since for $0 \leq z \leq 1$ the inverse tangent integrals have a series represantation of:

$${\rm Ti}_m (z)= \sum_{k=0}^{\infty} \frac{(-1)^k z^{2k+1}}{\left ( 2k+1 \right )^m}$$

[Tolaso J Kos]

Applications:

Setting $\theta=\frac{\pi}{2}$ back at the Clausen Integrals we have that:

\begin{equation} \int_{0}^{\pi/2} \ln \sin x \, {\rm d}x = - \frac{1}{2}{\rm Cl}_2 \left ( 2 \cdot \frac{\pi}{2} \right ) - \frac{\pi \ln 2}{2} \end{equation}

Now,

\begin{align*}
{\rm Cl}_2 \left ( \pi \right ) &=\sum_{n=1}^{\infty} \frac{\sin n \pi}{n^2} \\
&=0
\end{align*}

since $\sin n \pi=0$ forall $n \in \mathbb{N}$.

Thus returning back at $(1)$ we have verified the very well known result $\int_0^{\pi/2} \ln \sin x \, {\rm d}x = -\frac{\pi \ln 2}{2}$.

For $x=\frac{1}{4}$ we have that:

\begin{equation} \int_{0}^{\pi/4} \ln \sin x \, {\rm d}x = - \frac{1}{2}{\rm Cl}_2 \left ( 2 \cdot \frac{\pi}{4} \right ) - \frac{\pi \log 2}{4} \end{equation}

On the other hand:

\begin{align*}
{\rm Cl}_2 \left ( \frac{\pi}{2} \right ) &= \sum_{n=1}^{\infty} \frac{\sin \frac{n \pi}{2}}{n^2} \\
&= \cancelto{0}{\sum_{n=1}^{\infty} \frac{\sin \frac{2n \pi}{2}}{4n^2}} + \sum_{n=0}^{\infty} \frac{\sin \left ( \frac{(2n+1)\pi}{2} \right )}{\left ( 2n+1 \right )^2} \\
&= \sum_{n=0}^{\infty} \frac{\sin \left ( \frac{(2n+1)\pi}{2} \right )}{\left ( 2n+1 \right )^2} \\
&= \sum_{n=0}^{\infty} \frac{\sin \left ( n \pi + \frac{\pi}{2} \right )}{\left ( 2n+1 \right )^2} \\
&= \sum_{n=0}^{\infty} \frac{\cos n \pi}{\left ( 2n+1 \right )^2} \\
&= \sum_{n=0}^{\infty} \frac{(-1)^n}{\left ( 2n+1 \right )^2}\\
&=\mathcal{G}
\end{align*}

Thus $(2)$ returns the marvellous result.

$$\int_{0}^{\pi/4} \ln \sin x \, {\rm d}x = - \frac{1}{2}\mathcal{G} - \frac{\pi \ln 2}{4}$$

where $\mathcal{G}$ stands for the Catalan constant.

Similarly, we have that:

$$\int_{0}^{\pi/2} \ln \cos x \, {\rm d}x= - \frac{\pi \ln 2}{2} \quad , \quad \int_{0}^{\pi/4} \ln \cos x \, {\rm d}x = \frac{\mathcal{G}}{2} - \frac{\pi \ln 2}{4}$$

[Tolaso J Kos]

To complete the TUT [tutorial] here is some more advanced stuff on the Clausen function.

Duplication formula of even order: $\displaystyle {\rm Cl}_{2m}\left ( 2\theta \right )= 2^{2m-1} \left [ {\rm Cl}_{2m } (\theta) - {\rm Cl}_{2m}\left ( \pi - \theta \right ) \right ]$.

Proof: Indeed we have that:

\begin{align*}
{\rm Cl}_{2m } (\theta) - {\rm Cl}_{2m}\left ( \pi - \theta \right ) &= \sum_{n=1}^{\infty} \frac{\sin n \theta}{n^{2m}} - \sum_{n=1}^{\infty} \frac{\sin n(\pi-\theta)}{n^{2m}} \\
&= \sum_{n=1}^{\infty} \frac{\sin n \theta}{n^{2m}} - \sum_{n=1}^{\infty} \frac{\sin n \pi \cos n \theta -\cos n \pi \sin n \theta}{ n^{2m}}\\
&= \sum_{n=1}^{\infty} \frac{\sin n \theta}{n^{2m}} + \sum_{n=1}^{\infty} \frac{(-1)^n \sin n \theta}{n^{2m}}\\
&= 2 \sum_{n=1}^{\infty} \frac{\sin 2n \theta}{(2n)^{2m}} \\
&= \frac{1}{2^{2m-1}} {\rm Cl}_{2m} (2\theta)
\end{align*}

Duplication formula of odd order: $\displaystyle \text{Cl}_{2m+1}(2\theta)=\zeta(2m+1)+2^{2m}\left[\text{Cl}_{2m+1}(\theta)-\text{Cl}_{2m+1}(\pi-\theta)\right]$

Proof: Integrating the duplication formula of even order we have the result.

Applications:

• From the first formula (duplication formula of even order) if we set $\theta=\frac{\pi}{2}$ we get the following closed form:
  
  $${\rm Cl}_{2m} \left ( \frac{\pi}{2} \right )= \beta(2m)$$
  
  where $\beta$ stands for the Beta Dirichlet function.
• Applying the same formula to itself we get that:
  
  $$\text{Cl}_{2m}(4\theta)=4^{2m-1}\left[\text{Cl}_{2m}(\theta)-\text{Cl}_{2m}(\pi-\theta) -\text{Cl}_{2m}\left(\frac{\pi}{2}-\theta\right)+\text{Cl}_{2m}\left(\frac{\pi}{2}+\theta\right)\right]$$
• Applying the other duplication formula to itself we get that:
  
  $$\text{Cl}_{2m+1}(4\theta)=\zeta(2m+1)+4^{2m}\left[\text{Cl}_{2m+1}(\theta)-\text{Cl}_{2m+1}(\pi-\theta) -\text{Cl}_{2m+1}\left(\frac{\pi}{2}-\theta\right)+\text{Cl}_{2m+1}\left(\frac{\pi}{2}+\theta\right)\right]$$

Other interesting results (extracted only by basic use of trigonometry) are:

$$\sum_{k=1}^{\infty}\frac{\sin^2 n\theta}{n^{2m+1}}=\frac{1}{2}\Big[\zeta(2m+1)-\text{Cl}_{2m+1}(2\theta)\Big] \quad , \quad \sum_{n=1}^{\infty}\frac{\cos^2 n\theta}{n^{2m+1}}=\frac{1}{2}\Big[\zeta(2m+1)+\text{Cl}_{2m+1}(2\theta)\Big]$$

because

\begin{align*}
\sum_{n=1}^{\infty} \frac{\sin^2 n\theta}{n^{2m+1}} + \sum_{n=1}^{\infty} \frac{\cos^2 n \theta}{n^{2m+1}}&= \sum_{n=1}^{\infty} \frac{\cos^2 n\theta +\sin^2 n\theta}{n^{2m+1}} \\
&= \sum_{n=1}^{\infty} \frac{1}{n^{2m+1}}\\
&= \zeta(2m+1)
\end{align*}

and

\begin{align*}
\sum_{n=1}^{\infty} \frac{\sin^2 n\theta}{n^{2m+1}} - \sum_{n=1}^{\infty} \frac{\cos^2 n \theta}{n^{2m+1}} &= \sum_{n=1}^{\infty} \frac{\sin^2 n \theta - \cos^2 n \theta}{n^{2m+1}}\\
&= - \sum_{n=1}^{\infty} \frac{\cos 2n \theta}{n^{2m+1}}\\
&= - {\rm Cl}_{2m+1} (2\theta)
\end{align*}