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Logarithmo-trigonometric integrals


Post Wed Sep 11, 2013 3:19 pm

Posts: 138
Location: North Londinium, UK
Type 5:

\(\displaystyle \mathcal{Lc}_{\pm}(\theta)=\int_0^{\theta}\log(1\pm \cos x)\,dx\)





Proof:

For

\(\displaystyle \mathcal{Lc}_{+}(\theta)=\int_0^{\theta}\log(1+ \cos x)\,dx\,\)

use

\(\displaystyle \cos \frac{x}{2}=\left(\frac{1+\cos x}{2}\right)^{1/2}\,\) for \(\displaystyle |\theta|\le \pi\)

to obtain


\(\displaystyle \mathcal{Lc}_{+}(\theta)=\int_0^{\theta}\log\left(2\cos^2 \frac{x}{2}\right)\,dx=2\int_0^{\theta}\log\left(2\cos \frac{x}{2}\right)\,dx-\log 2\int_0^{\theta}\,dx=\)

\(\displaystyle -2\int_{\pi}^{\pi-\theta}\log\left(2\sin \frac{x}{2}\right)\,dx-\theta\log 2=\)

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


\(\displaystyle \therefore {Lc}_{+}(\theta)=2\text{Cl}_2(\pi-\theta)-\theta\log 2\)


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


Similarly, for

\(\displaystyle \mathcal{Lc}_{-}(\theta)=\int_0^{\theta}\log(1- \cos x)\,dx\)

use

\(\displaystyle \sin \frac{x}{2}=\left(\frac{1-\cos x}{2}\right)^{1/2}\,\) for \(\displaystyle 0\le \theta\le 2\pi\)

to obtain

\(\displaystyle \mathcal{Lc}_{-}(\theta)=\int_0^{\theta}\log\left(2\sin^2 \frac{x}{2}\right)\,dx=2\int_0^{\theta}\log\left(2\sin \frac{x}{2}\right)\,dx-\log 2\int_0^{\theta}\,dx=\)

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

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





\(\displaystyle \mathcal{Lc}_{+}(\theta)=\int_0^{\theta}\log(1+ \cos x)\,dx=2\text{Cl}_2(\pi-\theta)-\theta\log 2.\,\Box\)

\(\displaystyle \mathcal{Lc}_{-}(\theta)=\int_0^{\theta}\log(1- \cos x)\,dx=-2\text{Cl}_2(\theta)-\theta\log 2.\,\Box\)
Last edited by DreamWeaver on Wed Sep 11, 2013 7:56 pm, edited 1 time in total.

Post Wed Sep 11, 2013 3:22 pm

Posts: 138
Location: North Londinium, UK
Setting \(\displaystyle \theta = \pi/2\,\) in those last two eqns gives the classic results


\(\displaystyle \int_0^{\pi/2}\log(1+ \cos x)\,dx=2G-\frac{\pi}{2}\log 2\)

\(\displaystyle \int_0^{\pi/2}\log(1- \cos x)\,dx=-2G-\frac{\pi}{2}\log 2\)

Post Wed Sep 11, 2013 4:03 pm

Posts: 138
Location: North Londinium, UK
Type 6:


\(\displaystyle \int_0^{\theta}\log|\cos x + \sin x|\,dx\)



Proof:


\(\displaystyle \int_0^{\theta}\log|\cos x + \sin x|\,dx=\int_0^{\theta}\log\,\left|\sqrt{2}\left(\sin\frac{\pi}{4}\cos x + \cos\frac{\pi}{4}\sin x\right)\right|\,dx=\)

\(\displaystyle -\frac{\theta}{2}\log 2+\int_0^{\theta}\log\,\left|2\sin\left(x+\frac{\pi}{4}\right)\right|\,dx=\)

\(\displaystyle -\frac{\theta}{2}\log 2+\frac{1}{2}\int_{\pi/2}^{2\theta+\pi/2}\log\,\left|2\sin\frac{y}{2}\right|\,dy=\)

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

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



Thus

\(\displaystyle \int_0^{\theta}\log|\cos x + \sin x|\,dx=-\frac{\theta}{2}\log 2-\frac{1}{2}\text{Cl}_2\left(\frac{\pi}{2}+2\theta\right)+\frac{G}{2}.\,\Box\)
Last edited by DreamWeaver on Wed Sep 11, 2013 7:56 pm, edited 1 time in total.

Post Wed Sep 11, 2013 4:57 pm

Posts: 138
Location: North Londinium, UK
Type 7:

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



Proof:


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

\(\displaystyle \frac{1}{4}\int_0^{2\theta}y\,\log\left(2\sin \frac{y}{2}\right)\,dy-\frac{\theta^2}{2}\log 2=\)

\(\displaystyle \frac{1}{4}\int_0^{2\theta}x\,\left[-\frac{d}{dx}\text{Cl}_2(x)\right]\,dx-\frac{\theta^2}{2}\log 2=\)

\(\displaystyle -\frac{1}{4}x\,\text{Cl}_2(x)\biggr|_0^{2\theta}+\frac{1}{4}\int_0^{2\theta}\text{Cl}_2(x)\,dx-\frac{\theta^2}{2}\log 2\)

\(\displaystyle -\frac{\theta}{2}\,\text{Cl}_2(2\theta)-\frac{\theta^2}{2}\log 2+\frac{1}{4}\int_0^{2\theta}\text{Cl}_2(x)\,dx=\)

\(\displaystyle -\frac{\theta}{2}\,\text{Cl}_2(2\theta)-\frac{\theta^2}{2}\log 2+\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^2}\int_0^{2\theta}\sin kx\,dx=\)

\(\displaystyle -\frac{\theta}{2}\,\text{Cl}_2(2\theta)-\frac{\theta^2}{2}\log 2-\frac{1}{4}\sum_{k=1}^{\infty}\frac{\cos kx}{k^3}\Biggr|_0^{2\theta}=\)

\(\displaystyle -\frac{\theta}{2}\,\text{Cl}_2(2\theta)-\frac{\theta^2}{2}\log 2-\frac{1}{4}\sum_{k=1}^{\infty}\frac{\cos 2k\theta}{k^3}+\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^3}=\)

\(\displaystyle -\frac{\theta}{2}\,\text{Cl}_2(2\theta)-\frac{\theta^2}{2}\log 2-\frac{1}{4}\text{Cl}_3(2\theta)+\frac{1}{4}\zeta(3)\)


since

\(\displaystyle \text{Cl}_{2m+1}(\varphi)=\sum_{k=1}^{\infty}\frac{\cos k\varphi}{k^{2m+1}}\)





Thus

\(\displaystyle \int_0^{\theta}x\,\log(\sin x)\,dx=\frac{1}{4}\left[\zeta(3)-2\theta^2\log2-2\theta\text{Cl}_2(2\theta)-\text{Cl}_3(2\theta)\right].\,\Box\)
Last edited by DreamWeaver on Wed Sep 11, 2013 7:57 pm, edited 1 time in total.

Post Wed Sep 11, 2013 5:56 pm

Posts: 138
Location: North Londinium, UK
Type 8:

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




Proof:

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

\(\displaystyle \int_0^{\theta}x\log(\cos x)\,dx=-\int_{\pi/2}^{\pi/2-\theta}\left(y-\frac{\pi}{2}\right)\log(\sin y)\,dy=\)

\(\displaystyle -\int_{\pi/2}^{\pi/2-\theta}x\,\log(\sin x)\,dx+\frac{\pi}{2}\int_{\pi/2}^{\pi/2-\theta}\log(\sin x)\,dx=\)

Which, in terms of the solutions to Problems 4 & 7, is:


\(\displaystyle -\frac{1}{4}\Big[\zeta(3)-2\theta^2\log2-2\theta\text{Cl}_2(2\theta)-\text{Cl}_3(2\theta)\Big]_{\pi/2}^{\pi/2-\theta}
+\frac{\pi}{2}\Big[-\frac{1}{2}\text{Cl}_2(2\theta)-\theta\log 2\Big]_{\pi/2}^{\pi/2-\theta}\)


Which - unless I've made some hideously daft mistake - boils down to

\(\displaystyle \frac{\theta^2}{2}\log 2-\frac{\theta}{2}\text{Cl}_2(\pi-2\theta)+\frac{1}{4}\text{Cl}_3(\pi-2\theta)-\frac{1}{4}\text{Cl}_3(\pi)\)


And yet,

\(\displaystyle \text{Cl}_3(\pi)=\sum_{k=1}^{\infty}\frac{\cos \pi k}{k^3}=\sum_{k=1}^{\infty}\frac{(-1)^k}{k^3}=-\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^3}=-\eta(3)=\)

\(\displaystyle -\left(1+\frac{1}{2^3}-\frac{1}{3^3}+\frac{1}{4^3}-\cdots\right)=-\zeta(3)\left(1-\frac{2}{2^3}\right)=-\frac{3}{4}\zeta(3)\)






The proof is now complete, and the final result is

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

\(\displaystyle \frac{3}{16}\zeta(3)+\frac{\theta^2}{2}\log 2-\frac{\theta}{2}\text{Cl}_2(\pi-2\theta)+\frac{1}{4}\text{Cl}_3(\pi-2\theta).\,\Box\)




N.B. I will come back and double check all of these to makes sure I eliminate any silly mistakes... ;)
Last edited by DreamWeaver on Wed Sep 11, 2013 7:57 pm, edited 1 time in total.

Post Wed Sep 11, 2013 6:03 pm
zaidalyafey Global Moderator
Global Moderator

Posts: 357
This seems like a tutorial , you can add it to the tutorials section .
Wanna learn what we discuss , see Book

Post Wed Sep 11, 2013 6:22 pm

Posts: 138
Location: North Londinium, UK
zaidalyafey wrote:
This seems like a tutorial , you can add it to the tutorials section .


You're quite right there, Z, and that's kind of what I was thinking to start with... But I don't seem to have the option to start a new thread in the tutorials board. Can you help? Please and thank you! :D
Last edited by DreamWeaver on Wed Sep 11, 2013 7:06 pm, edited 1 time in total.

Post Wed Sep 11, 2013 7:04 pm

Posts: 138
Location: North Londinium, UK
Type 9:

\(\displaystyle \mathcal{I}(a\,;\, b)=\int_0^{\theta}\log(a\cos x +b\sin x)\,dx\)



Proof:

For non-zero a and b, and within the first quadrant \(\displaystyle 0\le\theta\le \pi/2\,\), the integrand is always positive, and hence converges. And so, to solve integrals of this type, we appeal to the auxiliary angle theorems of trigonometry, namely:

(1)\(\displaystyle \,\,\,\,\,\,\,\,\,\,a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x\pm\varphi);\,\,\,\,\, \varphi=\tan^{-1}(\mp b/a)\)

(2)\(\displaystyle \,\,\,\,\,\,\,\,\,\,a\cos x+b\sin x=\sqrt{a^2+b^2}\sin(x\pm\varphi);\,\,\,\,\, \varphi=\tan^{-1}(\pm a/b)\)


Thus

\(\displaystyle \mathcal{I}(a\,;\, b)=\int_0^{\theta}\log(a\cos x +b\sin x)\,dx\)

can be expressed as, say,


\(\displaystyle \frac{\theta}{2}\log(a^2+b^2)+\int_0^{\theta}\log\left[\sin(x-\varphi)\right]\,dx\)

for

\(\displaystyle \varphi=\tan^{-1}(- a/b)=-\tan^{-1}(a/b)\)



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


\(\displaystyle \int_0^{\theta}\log\left[\sin(x-\varphi)\right]\,dx\)

is then evaluated as follows

\(\displaystyle \int_0^{\theta}\log\left[\sin(x-\varphi)\right]\,dx=\int_0^{\theta}\log\left[2\sin(x-\varphi)\right]\,dx-\theta\log 2\)

Now substitute \(\displaystyle x-\varphi=y/2\,\Rightarrow\)

\(\displaystyle \frac{1}{2}\int_{-2\varphi}^{2\theta-2\varphi}\log\left(2\sin\frac{y}{2}\right)\,dy-\theta\log 2=\frac{1}{2}\left[\text{Cl}_2(-2\varphi)-\text{Cl}_2(2\theta-2\varphi)\right]-\theta\log 2\)


However, the Clausen function of order 2 is an odd function (a property it inherits from the sine function contained therein), hence

\(\displaystyle \text{Cl}_2(-\phi)=\sum_{k=1}^{\infty}\frac{\sin (-k\phi)}{k^2}=-\sum_{k=1}^{\infty}\frac{\sin (k\phi)}{k^2}=-\text{Cl}_2(\phi)\)



Thus

\(\displaystyle \int_0^{\theta}\log\left[\sin(x-\varphi)\right]\,dx=-\frac{1}{2}\left[\text{Cl}_2(2\varphi)+\text{Cl}_2(2\theta-2\varphi)\right]-\theta\log 2\)

and the final evaluation is


\(\displaystyle \int_0^{\theta}\log(a\cos x +b\sin x)\,dx=\)

\(\displaystyle \frac{\theta}{2}\log(a^2+b^2)-\frac{1}{2}\left[\text{Cl}_2(2\varphi)+\text{Cl}_2(2\theta-2\varphi)\right]-\theta\log 2.\,\Box\)
Last edited by DreamWeaver on Wed Sep 11, 2013 7:58 pm, edited 1 time in total.

Post Wed Sep 11, 2013 7:51 pm

Posts: 138
Location: North Londinium, UK
Type 10:

\(\displaystyle I_0=\int_0^1\log\Gamma(x)\,dx\)

Not strictly a log-trig integral this one, but a classic Gamma function integral that involves log-trig integration in the solution:





Proof:


Start with Euler's reflection formula for the Gamma Function:

\(\displaystyle \Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x}\)


Take the logarithm of both sides, and then integrate from 0 to 1 to obtain

\(\displaystyle \int_0^1\log\Gamma(x)\,dx+\int_0^1\log\Gamma(1-x)\,dx=\log\pi-\int_0^1\log(\sin \pi x)\,dx\)


Substitute \(\displaystyle x\to 1-y\,\) in the second loggamma integral to get:

\(\displaystyle 2\int_0^1\log\Gamma(x)\,dx=2\,I_0=\log\pi-\int_0^1\log(\sin \pi x)\,dx\)


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

\(\displaystyle \int_0^1\log(\sin \pi x)\,dx=\int_0^1\log(2\sin \pi x)\,dx-\log 2\int_0^1\,dx=\)

\(\displaystyle \int_0^1\log(2\sin \pi x)\,dx-\log 2\)


Next, set \(\displaystyle \pi x=y/2\,\) in the logsine integral to obtain

\(\displaystyle \int_0^1\log(2\sin \pi x)\,dx=\frac{1}{2\pi}\int_0^{2\pi}\log\left(2\sin\frac{y}{2}\right)\,dx=\)

\(\displaystyle -\frac{1}{2\pi}\text{Cl}_2(2\pi)=-\frac{1}{2\pi}\sum_{k=1}^{\infty}\frac{\sin 2\pi k}{k^2}=0\)




Hence

\(\displaystyle 2\int_0^1\log\Gamma(x)\,dx=\log\pi +\log 2\)




Or

\(\displaystyle \int_0^1\log\Gamma(x)\,dx=\log\sqrt{2\pi}.\,\Box\)

Post Wed Sep 11, 2013 8:13 pm

Posts: 138
Location: North Londinium, UK
For integrals similar to the last one - although far more advanced - I'd highly recommend the following paper:

http://arxiv.org/pdf/math/0012078.pdf



Here's but one example of the many, many beautiful integrals in the aforementioned paper (due to Olivier Espinosa and Victor H. Moll):

\(\displaystyle \int_0^1\log^2\Gamma(x)\,dx=\)

\(\displaystyle \frac{\gamma^2}{12}+\frac{\pi^2}{48}+\frac{\gamma}{3}\log\sqrt{2\pi}+\frac{4}{3}(\log\sqrt{2\pi})^2-(\gamma+2\log\sqrt{2\pi})\frac{\zeta'(2)}{\pi^2}+\frac{\zeta''(2)}{2\pi^2}\)




:shock: :shock: :shock: :shock: :shock:

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