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Integration Contest - Season 3

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Post Sun Sep 01, 2013 3:15 am
Shobhit Site Admin
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Problem 48

\(\displaystyle \int_0^{\frac{1}{2}}\frac{\log^{2} \left(x+\sqrt{1+x^2} \right)}{x}dx=\frac{\zeta(3)}{10}\)

Post Sun Sep 01, 2013 10:53 am
galactus User avatar
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\(\displaystyle \int_{0}^{1/2}\frac{log^{2}(x+\sqrt{x^{2}+1})}{x}dx\)




One way to tackle this one is by noting that \(\displaystyle log^{2}(x+\sqrt{x^{2}+1})=\sinh^{-1}(x)\)

Giving:

\(\displaystyle \int_{0}^{1/2}\frac{(\sinh^{-1}(x))^{2}}{x}dx\)

This has a series that can be derived by using the famous series for arcsin, \(\displaystyle (\sin^{-1}(x))^{2}=\frac{1}{2}\sum_{k=1}^{\infty}\frac{(2x)^{2k}}{k^{2}\binom{2k}{k}}\), and letting x=it or 'something or a ruther.' :)

Anyway, it is \(\displaystyle \int_{0}^{x}\frac{(\sinh^{-1}(t))^{2}}{t}dt=\frac{1}{4}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}(2x)^{2k}}{k^{3}\binom{2k}{k}}\)

Letting \(\displaystyle x=1/2\) gives the series:

\(\displaystyle \frac{1}{4}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{3}\binom{2k}{k}}\)

This is a rather famous series related to Apery's constant, \(\displaystyle \zeta(3)\), and is equal to \(\displaystyle \frac{2}{5}\zeta(3)\)

So, \(\displaystyle \frac{1}{4}\cdot \frac{2}{5}\zeta(3)=\frac{1}{10}\zeta(3)\)

Post Sun Sep 01, 2013 1:38 pm
Shobhit Site Admin
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An alternate approach is to prove that

\(\displaystyle \begin{align*} \int \frac{\text{arsinh}^2(x)}{x}dx &=-\frac{\text{arsinh}^3(x)}{3}+\text{arsinh}^2(x) \log \left(1-\left(x+\sqrt{1+x^2} \right)^2 \right) \\ &\quad +\text{arsinh}(x)\text{Li}_2 \left( \left(x+\sqrt{1+x^2} \right)^2 \right)-\frac{1}{2}\text{Li}_3 \left( \left(x+\sqrt{1+x^2} \right)^2 \right)

\end{align*}\)

Post Sun Sep 01, 2013 3:07 pm
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Cody


Maybe I'm overlooking something, but your evaluation seems circular. Can you evaluate that sum without evaluating that integral?

Post Sun Sep 01, 2013 3:38 pm
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Circular?. I didn't think it looks circular.

Though, it does rely on the already established sum \(\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{3}\binom{2k}{k}}=2/5\zeta(3)\)

I reckon there is yet another way by using \(\displaystyle \int_{0}^{log(\phi)}t^{2}\coth(t)dt=\frac{\zeta(3)}{10}, \;\ \phi=\frac{\sqrt{5}+1}{2}\)

By letting \(\displaystyle t=\sinh^{-1}(x)\).

Come to think of it, I think there is an Euler-like sum that could perhaps be used some how.

\(\displaystyle 4/25\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^{2}}H_{k}=\frac{1}{10}\zeta(3)\)

I have not tried it though.

Post Sun Sep 01, 2013 6:08 pm
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Using Cody's suggestion,


\(\displaystyle \int_{0}^{\log \phi} x^{2} \coth x \ dx = \int_{0}^{\log \phi} x^{2} \frac{1+e^{-2x}}{1-e^{-2x}} \ dx = \int_{0}^{\log \phi} x^{2} \Big( \frac{1}{1-e^{-2x}} + \frac{e^{-2x}}{1-e^{-2x}} \Big) \ dx\)

\(\displaystyle = \int_{0}^{\log \phi} x^{2} \Big( \sum_{k=0}^{\infty} e^{-2nx} + \sum_{k=1}^{\infty} e^{-2nx} \Big) \ dx = \int_{0}^{\log \phi} \Big(x^{2} + 2 x^{2} \sum_{n=1}^{\infty} e^{-2nx} \Big) \ dx\)

\(\displaystyle = \frac{\log^{3} \phi}{3} + 2 \sum_{n=1}^{\infty} \int_{0}^{\log \phi} x^{2} e^{-2nx} \ dx\)

\(\displaystyle = \frac{\log^{3} \phi}{3} - \frac{1}{2} \sum_{n=1}^{\infty} \frac{e^{-2nt} (2n^{2}t^{2}+2nt+1)}{n^{3}} \Big|^{\log \phi}_{0}\)

\(\displaystyle = \frac{\log^{3} \phi}{3} - t^{2} \sum_{n=1}^{\infty} \frac{e^{-2nt}}{n} - t \sum_{n=1}^{\infty} \frac{e^{-2nt}}{n^{2}} - \frac{1}{2} \sum_{n=1}^{\infty} \frac{e^{-2nt}}{n^{3}} \Big|^{\log \phi}_{0}\)

\(\displaystyle = \frac{\log^{3} \phi}{3} + t^{2} \ln(1-e^{-2t}) - t \ \text{Li}_{2}(e^{-2t})- \frac{1}{2} \text{Li}_{3} (e^{-2t}) \Big|^{\log \phi}_{0}\)

\(\displaystyle = \frac{\log^{3} \phi}{3} + \ln^{2} (\phi) \ln(1-\phi^{-2}) - \ln(\phi) \text{Li}_{2}(\phi^{-2}) - \frac{1}{2} \text{Li}_{3} (\phi^{-2}) + \frac{1}{2} \zeta(3)\)

\(\displaystyle = \frac{\log^{3} \phi}{3} + \ln^{2} (\phi) \ln(1-\phi^{-2}) - \ln (\phi) \Big( \frac{\pi^{2}}{15} - \ln^{2} (\phi) \Big) - \frac{1}{2} \Big( \frac{4}{5} \zeta(3) + \frac{2}{3} \ln^{3} (\phi) - \frac{2 \pi^{2}}{15} \ln (\phi) \Big) + \frac{1}{2} \zeta(3)\)

\(\displaystyle = \ln^{2}(\phi) \ln(1- \phi^{-2}) + \ln^{3}(\phi) + \frac{\zeta(3)}{10}\)

\(\displaystyle = \ln^{2}(\phi) \ln \left( \frac{1}{\phi} \right) + \ln^{3}(\phi) + \frac{\zeta(3)}{10} = - \ln^{3} (\phi) + \ln^{3}(\phi) + \frac{\zeta(3)}{10}\)

\(\displaystyle = \frac{\zeta(3)}{10}\)

Post Mon Sep 02, 2013 5:19 am
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

Good job, RV and C! :)

I posted the same integral a few months ago : http://integralsandseries.prophpbb.com/post371.html#p371

C had solved it using polylogarithms.

Problem 49

Prove that

\(\displaystyle \int_0^1 \frac{x^{p-\frac{1}{2}}}{(1-x)^p (1+qx)^p}dx = \frac{2}{\sqrt{\pi}}\Gamma \left(p+\frac{1}{2}\right) \Gamma(1-p) \cos^{2p}(\varphi) \frac{\sin \left( (2p-1)\varphi \right)}{(2p-1)\sin(\varphi)}\)

where \(\varphi = \arctan \sqrt{q}\)

This is from G&R's tables.

Problem 50

Evaluate

\(\displaystyle \int_{-\infty}^{\infty}\frac{e^{-px}}{\cosh(x)\cosh(x+a)\cosh(x+b)}dx\)

Post Mon Sep 02, 2013 8:56 am
galactus User avatar
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Very nice use of series, polylogs, etc. , G.

Post Mon Sep 02, 2013 11:21 am
Random Variable Integration Guru
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Using the integral representation of the Gaussian hypergeometric function,

\(\displaystyle \int_0^1 \frac{x^{p-\frac{1}{2}}}{(1-x)^p (1+qx)^p}dx = \frac{\Gamma(p+\frac{1}{2}) \Gamma(1-p)}{\Gamma(\frac{3}{2} )} {}_2F_{1} \Big( p,p+\frac{1}{2};\frac{3}{2};-q \Big) = \frac{2 \Gamma(p+\frac{1}{2}) \Gamma(1-p)}{\sqrt{\pi}} {}_2F_{1} \Big( p,p+\frac{1}{2};\frac{3}{2};-q \Big)\)

where \(\displaystyle = {}_2F_{1} \Big( p,p+\frac{1}{2};\frac{3}{2};-q \Big) = \sum_{k=0}^{\infty} \frac{\Gamma(p+k) \Gamma(p+\frac{1}{2}+k) \Gamma(\frac{3}{2})}{\Gamma(\frac{3}{2} + k) \Gamma(p) \Gamma(p+ \frac{1}{2})} \frac{(-q)^{k}}{k!}\)


I could write out the terms of the series, but I doubt it would be helpful.

Post Mon Sep 02, 2013 1:55 pm
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

The identity

\(\displaystyle _2F_1 \left( \frac{n+2}{2},\frac{n+1}{2};\frac{3}{2};-\tan^2 z\right)=\frac{\sin nz \cos^{n+1}z}{n\sin z}\)

fits here but I do not know it's proof. :cry:

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