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Logarithmo-trigonometric integrals


Post Wed Sep 11, 2013 9:19 pm
zaidalyafey Global Moderator
Global Moderator

Posts: 357
DreamWeaver wrote:
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



Well , that is strange . Shobhit must have to answer that . I can start the thread for you if you wish .
Wanna learn what we discuss , see Book

Post Wed Sep 11, 2013 9:33 pm

Posts: 138
Location: North Londinium, UK
OK, so maybe it's time for a trickier one... :twisted: :twisted: :twisted:

Type 11:


\(\displaystyle \mathbb{T}(m\,;\, n)=\int_{0}^{\pi/4}\log^n(\tan x)\log^m[-\log(\tan x)]\,dx\)


Which has the equivalent algebraic form

\(\displaystyle \mathbb{T}(m\,;\, n)=\int_0^1\frac{(\log x)^n\log^m(-\log x)}{1+x^2}\,dx\)




N.B. the following (excellent) article covers similar integrals - although Adamchik seems to have missed the far simpler proof I'll outline below...

http://www.cs.cmu.edu/~adamchik/articles/issac/issac97.pdf




Proof:


Actually, this one's not anywhere near as scary as it looks... Start with the hyperbolic integral

\(\displaystyle \mathcal{I}(p)=\int_0^{\infty}\frac{x^{p-1}}{\cosh x}\,dx=2\,\int_0^{\infty}\frac{x^{p-1}\,e^{-x}}{(1+e^{-2x})}\,dx=\)

\(\displaystyle 2\,\sum_{k=0}^{\infty}(-1)^k\int_0^{\infty}x^{p-1}\,e^{-(2k+1)x}\,dx\)

for the parameter \(\displaystyle p\in \mathbb{Q}^+\)



Make the substitution \(\displaystyle (2k+1)x=y\,\) to obtain

\(\displaystyle 2\,\sum_{k=0}^{\infty}(-1)^k\int_0^{\infty}x^{p-1}\,e^{-(2k+1)x}\,dx=\)

\(\displaystyle 2\,\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)}\int_0^{\infty}\left(\frac{y}{2k+1}\right)^{p-1}\,e^{-y}\,dy=\)

\(\displaystyle 2\,\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^p}\int_0^{\infty}x^{p-1}\,e^{-x}\,dy=2\,\Gamma(p)\,\beta(p)\)


Where \(\displaystyle \beta(x)\,\) is the Dirichlet Beta Function defined by

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

which has the special values

\(\displaystyle \beta(1)=\pi/4\)

\(\displaystyle \beta(2)=G\)

\(\displaystyle \beta(m)=\text{Ti}_m(1)\)



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


On the other hand, returning to

\(\displaystyle \mathcal{I}(p)=2\,\int_0^{\infty}\frac{x^{p-1}\,e^{-x}}{(1+e^{-2x})}\,dx=2\,\Gamma(p)\,\beta(p)\)

and setting \(\displaystyle y=e^{-x}\,\) yields

\(\displaystyle \int_0^{\infty}\frac{x^{p-1}\,e^{-x}}{(1+e^{-2x})}\,dx=\Gamma(p)\,\beta(p)=\int_0^1\frac{(-\log y)^{p-1}}{(1+y^2)}\,dy\)


Differentiating both sides m-times w.r.t. the parameter p gives

\(\displaystyle \sum_{j=0}^m\binom{m}{j}\Gamma^{(m-j)}(p)\,\beta^{(j)}(p)=\int_0^1\frac{(-\log y)^{p-1}\log^m(-\log y)}{(1+y^2)}\,dy=\)

\(\displaystyle \int_{0}^{\pi/4}\log^{p-1}(\tan x)\log^m[-\log(\tan x)]\,dx=\mathbb{T}(m\,;\, p-1)\)





And thus we obtain the final result, a binomial sum of derivatives of the Gamma and Dirichlet Beta Functions...


\(\displaystyle \mathbb{T}(m\,;\, n)=\int_{0}^{\pi/4}\log^n(\tan x)\log^m[-\log(\tan x)]\,dx=\)

\(\displaystyle \sum_{j=0}^m\binom{m}{j}\Gamma^{(m-j)}(n+1)\,\beta^{(j)}(n+1)\,.\Box\)
Last edited by DreamWeaver on Thu Sep 12, 2013 8:14 am, edited 1 time in total.

Post Wed Sep 11, 2013 9:36 pm

Posts: 138
Location: North Londinium, UK
zaidalyafey wrote:
DreamWeaver wrote:
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



Well , that is strange . Shobhit must have to answer that . I can start the thread for you if you wish .



Thanks! That's very kind of you... :D

It all depends on whether or not you think it'd be best to start a new thread, or just carry on with this one until Shobhit gets back, and have him move this 'ere thread...???

I'm easy either way :geek:

Post Wed Sep 11, 2013 9:43 pm

Posts: 138
Location: North Londinium, UK
On the subject of Type 11 - and with regards to the linked paper by Adamchik - using the methods outlined in said paper, I obtained an even more general form (also missed in that paper). Have a little read of it - if interested - and confirm for yourself that certain integrals of the form

\(\displaystyle \int (\tan x)^l\log^m(\tan x)\log^m(-\log(\tan x))\,dx\)

and

\(\displaystyle \int (\cot x)^l\log^m(\cot x)\log^m(-\log(\cot x))\,dx\)

are indeed possible... :o :o :o

Post Thu Sep 12, 2013 5:50 am
Shobhit Site Admin
Site Admin

Posts: 850
Location: Jaipur, India

gethin, I moved this thread to Tutorials. I think this should be better now. Also, now you can make new threads and posts in Tutorials Section.

Post Thu Sep 12, 2013 7:55 am

Posts: 138
Location: North Londinium, UK
Nice one, S! Thank you! :D It's very much appreciated...

Post Thu Sep 12, 2013 10:44 am

Posts: 138
Location: North Londinium, UK
One last point before moving on from Type 11... Apply the substitution \(\displaystyle x=\pi/2-y\,\) t0

\(\displaystyle \mathbb{T}(m\,;\, n)=\int_{0}^{\pi/4}\log^n(\tan x)\log^m[-\log(\tan x)]\,dx\)

to obtain

\(\displaystyle (-1)^n\int_{\pi/4}^{\pi/2}\log^n(\tan x)\log^m[\log(\tan x)]\,dx=\sum_{j=0}^m\binom{m}{j}\Gamma^{(m-j)}(n+1)\beta^{(j)}(n+1)\)



Case 1:


Vardi gave the beautiful, classic integral:

\(\displaystyle \int_{\pi/4}^{\pi/2}\log\log(\tan x)\,dx=\frac{\pi}{2}\log\left[\sqrt{2\pi}\frac{\Gamma(3/4)}{\Gamma(1/4)}\right]\)


By our generalized formula, this is \(\displaystyle \Gamma'(1)\beta(1)+\Gamma(1)\beta'(1)\,\). Now we work out the derivative of the Dirichlet Beta function at x=1. Using \(\displaystyle \beta(1)=\pi/4\,\), \(\displaystyle \Gamma'(1)=-\gamma\,\), and Vardi's result, we easily deduce that:

\(\displaystyle \beta'(1)=\sum_{k=0}^{\infty}(-1)^{k}\frac{\log(2k+1)}{(2k+1)}=\frac{\pi\gamma}{4}+\frac{\pi}{2}\log\left[\sqrt{2\pi}\frac{\Gamma(3/4)}{\Gamma(1/4)}\right]\)



Case 2:


Now let's find an evaluation of the second order Vardi-type integral:

\(\displaystyle \int_{\pi/4}^{\pi/2}\log^2[\log(\tan x)]\,dx\)


Our formula for Type 11 gives

\(\displaystyle \int_{\pi/4}^{\pi/2}\log^2[\log(\tan x)]\,dx=\Gamma''(1)\beta(1)+2\Gamma'(1)\beta'(1)+\Gamma(1)\beta''(1)\)


So we want the second order derivative of the Gamma function at x=1. This is easily found in terms of the digamma and trigamma functions:

\(\displaystyle \psi_1(x)=\frac{d^2}{dx^2}\log\Gamma(x)=\frac{d}{dx}\frac{\Gamma'(x)}{\Gamma(x)}=\frac{\Gamma(x)\Gamma''(x)-\Gamma'(x)^2}{\Gamma(x)^2}=\frac{\Gamma''(x)}{\Gamma(x)}-\psi_0(x)^2\)

\(\displaystyle \therefore\Gamma''(1)=\Gamma(1)\left[\psi_1(1)+\psi_0(1)^2\right]=\psi_1(1)+\gamma^2=\zeta(2)+\gamma^2=\frac{\pi^2}{6}+\gamma^2\)





Plugging this - as well as our earlier evaluation of the first order beta derivative - into our Type 11 formula, we then obtain the the following beautiful result for the second order Vardi-type integral:


\(\displaystyle \int_{\pi/4}^{\pi/2}\log^2[\log(\tan x)]\,dx=\beta''(1)+\frac{\pi^3}{24}-\frac{\pi\gamma^2}{4}-\pi\gamma\log\left[\sqrt{2\pi}\frac{\Gamma(3/4)}{\Gamma(1/4)}\right]\)




Case 3:


Following the same method as above, a little further calculation yields the third order Vardi-type integral to be:


\(\displaystyle \int_{\pi/4}^{\pi/2}\log^3[\log(\tan x)]\,dx=\)

\(\displaystyle \beta'''(1)-3\gamma\,\beta''(1)+\frac{\pi\gamma^3}{2}-\frac{\pi}{2}\zeta(3) + \left(\frac{\pi^3}{4}+\frac{3\pi\gamma^2}{2}\right)\log\left[\sqrt{2\pi}\frac{\Gamma(3/4)}{\Gamma(1/4)}\right]\)



:o :o :o

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