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Integration Contest - Season 3

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Post Wed Sep 11, 2013 11:20 am

Posts: 138
Location: North Londinium, UK
galactus wrote:
\(\displaystyle \int_{0}^{1/2}\frac{log^{2}(x+\sqrt{x^{2}+1})}{x}dx\)

I worked out a generalized (parametric) closed form for integrals of this type a few years back. Let me see now...

EDIT: I'd forgotten just how long and laborious the proof is, so I'll sketch out how it's done, but not add the entire proof (it'll be enough to demonstrate everything that's required for the full result).

Part 1:


\(\displaystyle \mathcal{S}_m(z)=\int_0^z\frac{\log^m(x+\sqrt{1+x^2})}{x}\,dx=\int_0^z\frac{(\sinh^{-1}x)^m}{x}\,dx\)

for \(\displaystyle 0 < z\le 1\)

Make the sub \(\displaystyle y=x+\sqrt{1+x^2}\,;\, x=\frac{y^2-1}{2y}\,;\, dx=\left(\frac{y^2+1}{2y^2}\right)\,dy\, \Rightarrow\)

\(\displaystyle \mathcal{S}_m(z)=\int_1^w\frac{(y^2+1)}{y\,(y^2-1)}(\log y)^m\,dy\)

where \(\displaystyle w=z+\sqrt{1+z^2}\)

Next, make the reciprocal sub \(\displaystyle y \to 1/x\) to get

\(\displaystyle (-1)^{m+1}\int_1^{1/w}\frac{(1+x^2)}{x\, (1-x^2)}\,(\log x)^m\,dx=\)

\(\displaystyle (-1)^{m+1}\int_1^{1/w}\frac{[(1-x^2)+x^2]}{x\, (1-x^2)}\,(\log x)^m\,dx+(-1)^{m+1}\int_1^{1/w}\frac{[1-(1-x^2)]}{x\, (1-x^2)}\,(\log x)^m\,dx\)


\(\displaystyle (-1)^{m+1}\mathcal{S}_m(z)=\)

\(\displaystyle \int_1^{1/w}\frac{(\log x)^m}{x}\,dx+\int_1^{1/w}\frac{x\,(\log x)^m}{(1-x^2)}\,dx+\)

\(\displaystyle \int_1^{1/w}\frac{(\log x)^m}{x\,(1-x^2)}\,dx-\int_1^{1/w}\frac{(\log x)^m}{x}\,dx=\)

\(\displaystyle \int_1^{1/w}\frac{x\,(\log x)^m}{(1-x^2)}\,dx+\int_1^{1/w}\frac{(\log x)^m}{x\,(1-x^2)}\,dx=\)

\(\displaystyle \frac{1}{2}\int_1^{1/w}\frac{x\,(\log x)^m}{(1-x)}\,dx+\frac{1}{2}\int_1^{1/w}\frac{x\,(\log x)^m}{(1+x)}\,dx+\)

\(\displaystyle \frac{1}{2}\int_1^{1/w}\frac{(\log x)^m}{x\,(1-x)}\,dx+\frac{1}{2}\int_1^{1/w}\frac{(\log x)^m}{x\,(1+x)}\,dx=\)

\(\displaystyle \frac{1}{2}\left(I_1+I_2+I_3+I_4\right)\,\) or

\(\displaystyle 2(-1)^{m+1}\mathcal{S}=\left(I_1+I_2+I_3+I_4\right)\)


My comp's being a tad slow today, so I'll post the proof in stages...

Post Wed Sep 11, 2013 12:21 pm

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

Solving \(\displaystyle I_1\,\)

\(\displaystyle I_1=\int_1^{1/w}\frac{x\,(\log x)^m}{(1-x)}\,dx=\int_1^{1/w}\frac{[1-(1-x)]\,(\log x)^m}{(1-x)}\,dx=\)

\(\displaystyle \int_1^{1/w}\frac{(\log x)^m}{(1-x)}\,dx-\int_1^{1/w}(\log x)^m\,dx\)

For this first integral, as well as the other three, we use the following classic calculus result (easily deduced by induction):

\(\displaystyle \int x^n(\log x)^m\,dx = m!\,x^{n+1}\,\sum_{j=0}^m(-1)^j\frac{(\log x)^{m-j}}{(m-j)!(n+1)^{j+1}}\)

\(\displaystyle \Rightarrow\)

\(\displaystyle I_1=\int_1^{1/w}\frac{(\log x)^m}{(1-x)}\,dx-\left[m!\,x\,\sum_{j=0}^m(-1)^j\frac{(\log x)^{m-j}}{(m-j)!}\right]_1^{1/w}=\)

\(\displaystyle \sum_{k=0}^{\infty}\int_1^{1/w}x^k(\log x)^m\,dx-\frac{(-1)^mm!}{w}\sum_{j=0}^m\frac{(\log w)^{m-j}}{(m-j)!}+(-1)^mm!=\)

\(\displaystyle \sum_{k=0}^{\infty}\left[m!\,x^{k+1}\,\sum_{j=0}^m(-1)^j\frac{(\log x)^{m-j}}{(m-j)!(k+1)^{j+1}}\right]_1^{1/w}-\)

\(\displaystyle \frac{(-1)^mm!}{w}\sum_{j=0}^m\frac{(\log w)^{m-j}}{(m-j)!}+(-1)^mm!=\)

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

\(\displaystyle \frac{(-1)^mm!}{w}\sum_{j=0}^m\frac{(\log w)^{m-j}}{(m-j)!}+(-1)^mm!=\)

\(\displaystyle (-1)^mm!\,\sum_{j=0}^m\frac{(\log w)^{m-j}}{(m-j)!}\,\text{Li}_{j+1}(1/w)+ (-1)^{m+1}m!\,\zeta(m+1)\)

\(\displaystyle \frac{(-1)^mm!}{w}\sum_{j=0}^m\frac{(\log w)^{m-j}}{(m-j)!}+(-1)^mm!=\)

\(\displaystyle (-1)^mm!\,\left[1-\zeta(m+1)+\sum_{j=0}^m\frac{(\log w)^{m-j}}{(m-j)!}\,\left[\text{Li}_{j+1}(1/w)+(1/w)\right]\right]\)

Apply the same type of approach to the three remaining (similar) integrals, change from w back to z, and you get a neat result that involving terms along the lines of

\(\displaystyle (\sinh^{-1}z)^{m-j}\,\text{Li}_{j+1}\left(\frac{1}{z+\sqrt{1+z^2}}\right)\)


\(\displaystyle (\sinh^{-1}z)^{m-j}\,\text{Li}_{j+1}\left(-\frac{1}{z+\sqrt{1+z^2}}\right)\)

Incidentally, although it's slightly trickier, the more general doubly-parametric form

\(\displaystyle \mathcal{S}_m(a;\,z) = \int_0^z\frac{(\sinh^{-1}(x/a))^m}{x}\,dx\)

has a similar closed form... :shock: :shock: :shock:

Post Wed Sep 11, 2013 12:40 pm

Posts: 138
Location: North Londinium, UK
On second thoughts, so I don't spoil the flow of this thread any more than I already have - :oops: :oops: :oops: - I'll work out the full result on paper, double check it, and then post the full proof on a new thread somewhere in the Integrals Board...

Post Wed Sep 11, 2013 4:24 pm
Random Variable Integration Guru
Integration Guru

Posts: 381
It would seem that what I did on the previous page to evaluate \(\displaystyle \int_{0}^{\frac{1}{2}} \frac{\text{arcsinh}^{2} x}{x} \ dx = \int^{\log \phi}_{0} x^{2} \coth x \ dx\) where \(\displaystyle \phi\) is the golden ratio could be extended to evaluate \(\displaystyle \int_{0}^{\theta} \frac{\text{arcsinh}^{2} x}{x} \ dx\).

\(\displaystyle \int_{0}^{\theta} \frac{\text{arcsinh}^{m} x}{x} \ dx = \int^{\log (\theta+ \sqrt{\theta^{2}+1})}_{0} x^{m} \coth x \ dx = \int_{0}^{\log (\theta+ \sqrt{\theta^{2}+1})} \Big(x^{m} + 2 x^{m} \sum_{n=1}^{\infty} e^{-2nx} \Big) \ dx\)

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

\(\displaystyle = \frac{\text{arcsinh}^{m+1} (\theta)}{m+1 } + 2 \sum_{n=1}^{\infty} \Bigg( -\big(\theta + \sqrt{\theta^{2}+1} \big)^{-2n} \sum_{j=1}^{m} \frac{\text{arcsinh}^{m-j+1}(\theta)}{(2n)^{j}} \frac{m!}{(m-j+1)!} + \frac{m!}{(2n)^{m+1}} \big( 1-(\theta + \sqrt{\theta^{2}+1} \big)^{-2n} \Bigg)\)

\(\displaystyle = \frac{\text{arcsinh}^{m+1} (\theta)}{m+1 }- 2 m! \sum_{j=1}^{m} \frac{\text{arcsinh}^{m-j+1}(\theta)}{2^{j}(m-j+1)!} \text{Li}_{j} \Bigg( \frac{1}{(\theta + \sqrt{\theta^{2}+1})^{2}} \Bigg) + \frac{m!}{2^{m}} \zeta(m+1) - \frac{m!}{2^{m}} \text{Li}_{m+1} \Bigg( \frac{1}{(\theta + \sqrt{\theta^{2}+1})^{2}} \Bigg)\)

Post Sat Sep 14, 2013 12:26 pm
galactus User avatar
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Shobhit wrote:
I think we should move on then:

Problem 44

\(\displaystyle \int_0^\infty \frac{1}{(1+x^2)\left( 1-a \cos \left( x-\frac{1}{x}\right)+a^2\right)}dx = \frac{\pi}{2}\frac{1}{1-a^2}\frac{e^2+a}{e^2-a}\)

\(\displaystyle a<1\)

I have not yet tried this one.

I tried running this integral through Maple and Mathematica and neither would give me any result whatsoever. Not even a numerical one for a check. So, I made a sub and hammered it into another form.

Sorry I have not been around. Too busy with work and other matters. Anyway, I plodded along with this one off and on as I had time. I think the condition should be \(\displaystyle |a|<1\).

I arrive at a solution different than the posted one. As is, if a = -1, the given solution is undefined. May I ask, S, where you found this?.

I made the obvious sub of \(\displaystyle t=x-\frac{1}{x}\), and arrived at \(\displaystyle I(t)=\int_{-\infty}^{\infty}\frac{1}{(t^{2}+4)(1-a\cos(t)+a^{2})}dt\)

Now, most integrals similar to this one have 2pi as the upper limit.

\(\displaystyle \sum_{n=-\infty}^{\infty}\int_{2\pi n}^{2\pi (n+1)}I(t)\)

\(\displaystyle =\int_{0}^{2\pi}\sum_{n=-\infty}^{\infty}I(2\pi n+t)\)

\(\displaystyle \int_{0}^{2\pi}\sum_{n=-\infty}^{\infty}\frac{1}{((2\pi n+t)^{2}+4)(1-a\cos(t)+a^{2})}dt\)

Use the unit circle as a contour and note that the sum evaluates to \(\displaystyle \sum_{n=-\infty}^{\infty}\frac{1}{(2\pi n+t)^{2}+4}=\frac{\sinh(2)}{4(\cosh(2)-\cos(t))}\).

It can be evaluated rather easily by using the typical \(\displaystyle \pi \cot(\pi z)\)

Resulting in

\(\displaystyle \frac{\sinh(2)}{4}\int_{0}^{2\pi}\frac{1}{(\cosh(2)-\cos(t))(1-a\cos(t)+a^{2})}dt\)

Use the classic \(\displaystyle z=e^{it}\) and get:

\(\displaystyle \int_{0}^{2\pi}\frac{4ie^{2}z}{(e^{2}z^{2}-(1+e^{4})z+e^{2})(2(a^{2}+1)z-a(z^{2}+1))}dz\)

Finding the residues at the respective poles. The left factor in the denominator has pole \(\displaystyle e^{-2}\) and the right has pole \(\displaystyle \frac{a^{2}+1-\sqrt{a^{4}+a^{2}+1}}{a}\) lying inside the contour.

Summing the residues and multiplying by \(\displaystyle \frac{2\pi i\sinh(2)}{4}\):

\(\displaystyle \frac{2\pi i\sinh(2)}{4}\left(\frac{-i}{\sinh(2)(a^{2}+1-a\cosh(2))}+\frac{ai}{(a^{2}+1-a\cosh(2))\sqrt{a^{4}+a^{2}+1}}\right)\)

The final result being \(\displaystyle \frac{\pi\sinh(2)}{2(a^{2}+1-a\cosh(2))}\left(\frac{1}{\sinh(2)}-\frac{a}{\sqrt{a^{4}+a^{2}+1}}\right)\)

Of course, this can probably be written in various forms. Maybe it can be equated to the given solution in some manner. I checked several values of 'a' for 0<a<1 and they worked. I also checked the general form with Maple and it checked.

Post Sat Sep 14, 2013 5:46 pm
Random Variable Integration Guru
Integration Guru

Posts: 381
That was smart to break up the integral like that.

That integral will diverge if \(\displaystyle 1 - a \cos t + a^{2}\) has zeroes on the real axis.

So it will diverge if \(\displaystyle t = \arccos \Big( \frac{1+a^{2}}{a} \Big)\) is real, that is if \(\displaystyle -1 \le \frac{a^{2}+1}{a} \le 1\).

But that inequality is not satisfied by any real value of \(\displaystyle a\) (contrary to what I initially said).

So the integral must converge for all real values of \(\displaystyle a\).

They must have had to restrict the values of \(\displaystyle a\) to do whatever it is they did.

Post Sat Sep 14, 2013 7:13 pm
galactus User avatar
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Hey RV:

Yeah, I figured there may be a typo in the original statement or some restrictions as you mentioned.

I tried all sorts of values of 'a' in the original integral and could not find any value that arrived at a solution. Except for a=0.

Which gives \(\displaystyle \frac{\pi}{2}\)

Maple and Mathematica both kept spitting back the integral. Which usually means there is no solution or it diverges. But, then again, maybe the math engines can simply not do it.

I went ahead and assumed the common restriction |a|<1, and tried for the 2 pi in the upper limit of integration because other integrals I have ran across that have the same Poisson-type factor in the denominator have this as their upper limit.

After getting the 2 pi, then it became more manageable.

As you mentioned RV, another thought was that by considering the solutions to \(\displaystyle \cos(t)=a+1/a\), and noting they are of the form \(\displaystyle bi\) where b is real and positive.

The other solutions are at \(\displaystyle b_{n}=bi+2\pi n\), and the residue there is \(\displaystyle \frac{1}{i((bi+2\pi n)^{2}+4)\sinh(b)}\)

This leads to an infinite series

Post Fri Sep 20, 2013 7:12 am
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

galactus, your solution is very impressive! :) I checked it for a few values of \(a\) and it seems to give correct results.

Post Sun Oct 19, 2014 3:54 am

Posts: 70
Location: Montreal, Canada

Note: I answered a question posted by Zaid Al-yafi but I do not know how it ended up on this page.

The integral can have the closed form

$$I = \displaystyle \int^{\infty}_0 e^{- \beta \, x}\left( \frac{1}{x}-\coth(x)\right)\, dx= \frac{1}{\beta} - \ln\left( \frac{\beta}{2}\,\right) + \psi \left( \frac{\beta}{2}\,\right),$$

where $ \psi(x) $ is the digamma function.
Last edited by mathematics on Sun Oct 19, 2014 4:01 am, edited 1 time in total.

Post Sun Oct 18, 2015 2:53 pm

Posts: 16
mathematics wrote:
Note: I answered a question posted by Zaid Al-yafi but I do not know how it ended up on this page.

The integral can have the closed form

$$I = \displaystyle \int^{\infty}_0 e^{- \beta \, x}\left( \frac{1}{x}-\coth(x)\right)\, dx= \frac{1}{\beta} - \ln\left( \frac{\beta}{2}\,\right) + \psi \left( \frac{\beta}{2}\,\right),$$

where $ \psi(x) $ is the digamma function.

$$I\left( \beta \right)=\int\limits_{0}^{+\infty }{e^{-\beta x}\left( \frac{1}{x}-\coth x \right)dx}=\int\limits_{0}^{+\infty }{e^{-\beta x}\left( \frac{1}{x}-\frac{1+e^{-2x}}{1-e^{-2x}} \right)dx}=\int\limits_{0}^{+\infty }{e^{-\frac{1}{2}\beta x}\left( \frac{1}{x}-\frac{1}{2}\cdot \frac{1+e^{-x}}{1-e^{-x}} \right)dx}$$

$$=\int\limits_{0}^{+\infty }{\left( \frac{e^{-\frac{1}{2}\beta x}}{x}-\frac{1}{2}\cdot \frac{e^{-\frac{1}{2}\beta x}+e^{-\frac{1}{2}\beta x-x}}{1-e^{-x}} \right)dx}$$

$$=\int\limits_{0}^{+\infty }{\left( \frac{e^{-x}+e^{-\frac{1}{2}\beta x}-e^{-x}}{x}-\frac{1}{2}\cdot \frac{e^{-\frac{1}{2}\beta x}+e^{-\frac{1}{2}\beta x}-e^{-\frac{1}{2}\beta x}+e^{-\frac{1}{2}\beta x-x}}{1-e^{-x}} \right)dx}$$

$$=\int\limits_{0}^{+\infty }{\left( \frac{e^{-x}}{x}-\frac{e^{-\frac{1}{2}\beta x}}{1-e^{-x}} \right)dx}+\frac{1}{2}\int\limits_{0}^{+\infty }{\left( \frac{e^{-\frac{1}{2}\beta x}-e^{-\frac{1}{2}\beta x-x}}{1-e^{-x}} \right)dx}+\int\limits_{0}^{+\infty }{\left( \frac{e^{-\frac{1}{2}\beta x}-e^{-x}}{x} \right)dx}$$

$$=\psi \left( \frac{\beta }{2} \right)-\ln \left( \frac{\beta }{2} \right)+\frac{1}{2}\int\limits_{0}^{+\infty }{e^{-\frac{1}{2}\beta x}dx}=\psi \left( \frac{\beta }{2} \right)-\ln \left( \frac{\beta }{2} \right)+\frac{1}{a}$$


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