Board index Tutorials Logarithmo-trigonometric integrals

## Logarithmo-trigonometric integrals

### Re: Logarithmo-trigonometric integrals

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.

### Re: Logarithmo-trigonometric integrals

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

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

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

### Re: Logarithmo-trigonometric integrals

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.

### Re: Logarithmo-trigonometric integrals

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.

### Re: Logarithmo-trigonometric integrals

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.

### Re: Logarithmo-trigonometric integrals

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

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

### Re: Logarithmo-trigonometric integrals

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!
Last edited by DreamWeaver on Wed Sep 11, 2013 7:06 pm, edited 1 time in total.

### Re: Logarithmo-trigonometric integrals

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.

### Re: Logarithmo-trigonometric integrals

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:

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

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

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

### Re: Logarithmo-trigonometric integrals

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):

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

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

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