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Logarithmo-trigonometric integrals


Post Mon Sep 09, 2013 9:53 am

Posts: 138
Location: North Londinium, UK
Hi all! :D

Definite integrals involving logarithms of trig functions are one of my main areas of interest, so I thought I'd pose a few problems on here, starting off with easy ones, and then gradually ramping up the difficulty level. Also, it goes with out saying that I'd be very interested in giving any integrals you guys and gals fancy posting a go. I'm a self-taught amateur, mind, and not on the same level as the rest of you - I can see that by perusing the forum - but I'll happily give anything a go... ;)

Problem 1:

Without residues - and without using Lewin's book! - evaluate

\(\displaystyle S_m(\theta)=\int_0^{\theta}x^m\log(\sin x)\,dx\)

in terms of Clausen functions and transcendental constants, where the Clausen functions are defined by:

\(\displaystyle \text{Cl}_2(\theta)=-\int_0^{\theta}2\log\bigg|\sin \frac{x}{2}\bigg|\,dx\)

\(\displaystyle \text{Cl}_n(\theta) = \left\{
\begin{array}{l l}
\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^n} & \quad \text{if $n$ is even}\\
\sum_{k=1}^{\infty}\frac{\cos k\theta}{k^n} & \quad \text{if $n$ is odd}
\end{array} \right.\)

\(\displaystyle \text{Sl}_n(\theta) = \left\{
\begin{array}{l l}
\sum_{k=1}^{\infty}\frac{\cos k\theta}{k^n} & \quad \text{if $n$ is even}\\
\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^n} & \quad \text{if $n$ is odd}
\end{array} \right.\)



Hint: at some point you might split the finite series into odd resp. even terms, using

\(\displaystyle \sum_{k=0}^m f(k)=\sum_{j=0}^{\lfloor m/2\rfloor}f(2j)+\sum_{j=0}^{\lfloor (m-1)/2\rfloor}f(2j+1)\)

Post Mon Sep 09, 2013 12:26 pm
Random Variable Integration Guru
Integration Guru

Posts: 381
This is not a closed form, but this is how I would evaluate the integral for specific values of \(\displaystyle m\).


Using the fact that \(\displaystyle \log(1-e^{2ix}) =\log(2 \sin x) + i \Big(x- \frac{\pi}{2} \Big)\) we have

\(\displaystyle \log 2 \int_{0}^{\theta} x^{m} dx + \int_{0}^{\theta} x^{m} \log (\sin x) \ dx = \int_{0}^{\theta} x^{m} \ln(1-e^{2ix}) \ dx - i \int_{0}^{\theta} x^{m} \Big(x- \frac{\pi}{2}\Big) \ dx\)


Then If we impose the restriction \(\displaystyle 0 < \theta < \pi\) ,

\(\displaystyle \int_{0}^{\theta} x^{m} \log (\sin x) \ dx = - \log 2 \int_{0}^{\theta} x^{m} \ dx + \text{Re} \int_{0}^{\theta} x^{m} \ln(1-e^{2ix}) \ dx\)

\(\displaystyle = - \log(2) \frac{\theta^{m+1}}{m+1} - \text{Re} \int_{0}^{\theta} x^{m} \sum_{n=1}^{\infty} \frac{e^{2inx}}{n} \ dx\)

\(\displaystyle = - \log(2) \frac{\theta^{m+1}}{m+1} - \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\theta} x^{m} \cos(2 nx) \ dx\)

Post Mon Sep 09, 2013 2:10 pm

Posts: 138
Location: North Londinium, UK
Quite right, RV... And you're just a few small steps away from the full closed form, since

\(\displaystyle \int x^{2n}\cos x\,dx=(2n)!\left[ \sum_{k=0}^{n}(-1)^k\frac{x^{2n-2k}}{(2n-2k)!}\sin x + \sum_{k=0}^{n-1}(-1)^k\frac{x^{2n-2k-1}}{(2n-2k-1)!}\cos x\right]\)

and

\(\displaystyle \int x^{2n+1}\cos x\,dx=(2n+1)!\left[ \sum_{k=0}^{n}(-1)^k\frac{x^{2n-2k+1}}{(2n-2k+1)!}\sin x + \sum_{k=0}^{n}(-1)^k\frac{x^{2n-2k}}{(2n-2k)!}\cos x\right]\)

Post Mon Sep 09, 2013 3:12 pm
Random Variable Integration Guru
Integration Guru

Posts: 381
Is there a way to derive that other than repeatedly integrating by parts?

Post Mon Sep 09, 2013 3:44 pm

Posts: 138
Location: North Londinium, UK
Not as far as I know, but if memory serves, you can prove it by induction.

Let me know if and when you want the full result, and I'll post a proof.



Gethin

Post Wed Sep 11, 2013 12:54 pm

Posts: 138
Location: North Londinium, UK
Let's post a few easy ones...


Type 1:

\(\displaystyle \int_0^{\pi/4}\log^m(\tan x)\, dx\)




Solution:

\(\displaystyle \int_0^{\pi/4}\log^m(\tan x)\, dx=\int_0^1\frac{(\log x)^m}{(1+x^2)}\,dx=\)

\(\displaystyle \sum_{k=0}^{\infty}(-1)^k\int_0^1x^{2k}(\log x)^m\,dx=(-1)^mm!\,\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^{m+1}}=(-1)^mm!\,\beta(m+1).\,\Box\)


Where

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

is the Dirichlet Beta Function.


The case x=2 is Catalan's constant:

\(\displaystyle \beta(2)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^2}=G\)
Last edited by DreamWeaver on Wed Sep 11, 2013 7:54 pm, edited 1 time in total.

Post Wed Sep 11, 2013 1:05 pm

Posts: 138
Location: North Londinium, UK
Type 2:

\(\displaystyle \int_0^{\pi/2}\log^m(\tan x)\, dx\)



Solution:


\(\displaystyle \int_0^{\pi/2}\log^m(\tan x)\, dx=\int_0^{\pi/4}\log^m(\tan x)\, dx+\int_{\pi/4}^{\pi/2}\log^m(\tan x)\, dx\)

Make the sub \(\displaystyle x=\pi/2 - y\,\) on the second integral to obtain

\(\displaystyle \int_0^{\pi/2}\log^m(\tan x)\, dx=\int_0^{\pi/4}\log^m(\tan x)\, dx+\int_0^{\pi/4}\log^m(\cot x)\, dx=\)

\(\displaystyle \int_0^{\pi/4}\log^m(\tan x)\, dx+(-1)^m\,\int_0^{\pi/4}\log^m(\tan x)\, dx=\)

\(\displaystyle [1+(-1)^m]\int_0^{\pi/4}\log^m(\tan x)\, dx\)




\(\displaystyle \therefore \int_0^{\pi/2}\log^m(\tan x)\, dx = [1+(-1)^m]\,m!\,\beta(m+1).\,\Box\)
Last edited by DreamWeaver on Wed Sep 11, 2013 7:54 pm, edited 1 time in total.

Post Wed Sep 11, 2013 1:54 pm

Posts: 138
Location: North Londinium, UK
Type 3:


\(\displaystyle \mathcal{T}_m(\theta)=\int_0^{\theta}\log^m(\tan x)\,dx\)



Solution:

\(\displaystyle \int_0^{\theta}\log^m(\tan x)\,dx=\int_0^{z=\tan\theta}\frac{(\log x)^m}{1+x^2}\,dx=\)

\(\displaystyle \sum_{k=0}^{\infty}(-1)^k\int_0^zx^{2k}(\log x)^m\,dx\)


Let \(\displaystyle x\to z\,y\,\) to get

\(\displaystyle \sum_{k=0}^{\infty}(-1)^k\int_0^zx^{2k}(\log x)^m\,dx=\sum_{k=0}^{\infty}(-1)^k\,z^{2k+1}\int_0^1y^{2k}(\log z+\log y)^m\,dy\)


Use the finite binomial theorem on the logarithmic part

\(\displaystyle \sum_{j=0}^m\binom{m}{j}(\log z)^{m-j}\sum_{k=0}^{\infty}(-1)^k\,z^{2k+1}\int_0^1y^{2k}(\log y)^j\,dy\)

and the classic - easily provable - result

\(\displaystyle \int_0^1x^n(\log x)^m\,dx=\frac{(-1)^m\,m!}{(n+1)^{m+1}}\)

to obtain

\(\displaystyle \sum_{j=0}^m(-1)^j\,j!\,\binom{m}{j}(\log z)^{m-j}\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)^{j+1}}=\)

\(\displaystyle m!\,\sum_{j=0}^m(-1)^j\,\frac{(\log z)^{m-j}}{(m-j)!}\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)^{j+1}}\)






Finally, observe that for \(\displaystyle 0\le z\le 1\,\) the Inverse Tangent Integrals

\(\displaystyle \text{Ti}_1(z)=\frac{\tan^{-1}z}{z}\)

\(\displaystyle \text{Ti}_{m+1}(z)=\int_0^z\frac{\text{Ti}_m(x)}{x}\)

have the series representation

\(\displaystyle \text{Ti}_m(z)=\sum_{k=0}^{\infty}(-1)^k\frac{z^{2k+1}}{(2k+1)^k}\)

and the proof is complete:




\(\displaystyle \mathcal{T}_m(\theta)=m!\,\sum_{j=0}^m(-1)^j\frac{\log^{m-j}(\tan \theta)}{(m-j)!}\,\text{Ti}_{j+1}(\tan \theta).\,\Box\)
Last edited by DreamWeaver on Wed Sep 11, 2013 7:55 pm, edited 1 time in total.

Post Wed Sep 11, 2013 2:32 pm

Posts: 138
Location: North Londinium, UK
Type 4:


\(\displaystyle \mathcal{S}(\theta)=\int_0^\theta\log(\sin x)\,dx\)

\(\displaystyle \mathcal{C}(\theta)=\int_0^\theta\log(\cos x)\,dx\)




Proof:

\(\displaystyle \mathcal{S}(\theta)=\int_0^{\theta}\log(\sin x)\,dx=\int_0^\theta\log(2\sin x)\,dx-\log 2\int_0^{\theta}\,dx=\)

\(\displaystyle \frac{1}{2}\int_0^{2\theta}\log[2\sin (y/2)]\,dx-\theta\log 2\)


In terms of the Clausen function

\(\displaystyle \text{Cl}_2(\phi)=-\int_0^{\phi}|\log(2\sin (x/2))|\,dx=\sum_{k=1}^{\infty}\frac{\sin k\phi}{k^2}\)


the answer is, therefore


\(\displaystyle \mathcal{S}(\theta)=-\frac{1}{2}\text{Cl}_2(2\theta)+\frac{1}{2}\text{Cl}_2(0)-\theta\log 2=-\frac{1}{2}\text{Cl}_2(2\theta)-\theta\log 2\)


------------------------


Next, make the sub \(\displaystyle x=\pi/2-y\,\) in \(\displaystyle \mathcal{C}(\theta)\,\) to get

\(\displaystyle \mathcal{C}(\theta)=-\int_{\pi/2}^{\pi/2-\theta}\log\left[\cos\left(\frac{\pi}{2}-y\right)\right]\,dy=-\int_{\pi/2}^{\pi/2-\theta}\log(\sin y)\,dy=\)

\(\displaystyle \mathcal{S}(\pi/2)-\mathcal{S}(\pi/2-\theta)=-\frac{1}{2}\text{Cl}_2(\pi)+\frac{1}{2}\text{Cl}_2(\pi-2\theta)-\theta\log 2=\)

\(\displaystyle \frac{1}{2}\text{Cl}_2(\pi-2\theta)-\theta\log 2\)


------------------------




The proof is complete:


\(\displaystyle \mathcal{S}(\theta)=\int_0^\theta\log(\sin x)\,dx=-\frac{1}{2}\text{Cl}_2(2\theta)-\theta\log 2.\,\Box\)

\(\displaystyle \mathcal{C}(\theta)=\int_0^\theta\log(\cos x)\,dx=\frac{1}{2}\text{Cl}_2(\pi-2\theta)-\theta\log 2.\,\Box\)
Last edited by DreamWeaver on Wed Sep 11, 2013 7:55 pm, edited 1 time in total.

Post Wed Sep 11, 2013 2:40 pm

Posts: 138
Location: North Londinium, UK
Using the previous two results

\(\displaystyle \mathcal{S}(\theta)=\int_0^\theta\log(\sin x)\,dx=-\frac{1}{2}\text{Cl}_2(2\theta)-\theta\log 2\)

\(\displaystyle \mathcal{C}(\theta)=\int_0^\theta\log(\cos x)\,dx=\frac{1}{2}\text{Cl}_2(\pi-2\theta)-\theta\log 2\)



setting \(\displaystyle \theta=\pi/4\,\), and then using the (easily proven) result \(\displaystyle \text{Cl}_2(\pi/2)=G\,\) (Catalan's constant), we obtain the classic results:


\(\displaystyle \int_0^{\pi/4}\log(\sin x)\,dx=-\frac{G}{2}-\frac{\pi}{4}\log 2\)

\(\displaystyle \int_0^{\pi/4}\log(\cos x)\,dx=\frac{G}{2}-\frac{\pi}{4}\log 2\)



This is exactly as we would expect, since


\(\displaystyle \int_0^{\pi/4}\log(\tan x)\,dx=-G=\int_0^{\pi/4}\log(\sin x)\,dx-\int_0^{\pi/4}\log(\cos x)\,dx\)

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