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## Integration Contest - Season 5

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### Re: Integration Contest - Season 5

Fri May 29, 2015 7:26 pm

Posts: 38
galactus wrote:
Here's another integral if anyone wants a go.

$$\int_{-\infty}^{\infty}\frac{x\ln(|x|)\ln(|x-1|)}{(x^{2}+1)(x^{2}+3)}dx=\frac{\pi \ln(8/9)}{24}$$

That should be $$\frac{\pi^2}{24}\log\frac{8}{9}$$

### Re: Integration Contest - Season 5

Fri May 29, 2015 7:40 pm
galactus
Global Moderator

Posts: 902
Yes. Thanks for the catch. Typo

### Re: Integration Contest - Season 5

Sat May 30, 2015 7:32 pm

Posts: 12

P.S. I think we need to make a new rule to not use complex analysis; it makes things quite easy
P.P.S. I still haven't managed to solve my (the previous) integral without invoking the Dirichlet Beta (which we still haven't proven). I very nearly proved beta'(0) (which would be sufficient for beta'(1)), until I realised the series I was using diverges (it looked so close to the correct answer though!)

### Re: Integration Contest - Season 5

Sat May 30, 2015 9:28 pm
galactus
Global Moderator

Posts: 902
Yep, I thought contours would make that a pushover. Try a real method if you wish.

I calculated the residues and saw it was coming together.

What exactly is it you want to prove regarding the Dirichlet Beta?.

The derivative of DB is related to the derivative of Zeta...as I kind of mention below. Here is a link. Is this what you mean?. http://engineeringandmathematics.blogsp ... ation.html

Adamchik wrote a paper dedicated to integrals of the type $log(log(u))$ and so on.

He even derived a general form:

$$\int_{0}^{1}\frac{u^{p-1}}{1+u^{n}}\log\log(1/u)du=\frac{\gamma+\log(2n)}{2n}\left[\psi(\frac{p}{2n})-\psi(\frac{n+p}{2n})\right]+\frac{1}{2n}\left[\zeta'(1,\frac{p}{2n})-\zeta'(1,\frac{n+p}{2n})\right]$$

for our integral, let $p=1, \;\ n=1$

note that $$\zeta'(1,1/4)-\zeta'(1,3/4)=\pi\left[\gamma+4\log(2)+3\log(\pi)-4\log\Gamma(1/4)\right]$$

and $$\psi(1/4)-\psi(3/4)=-\pi$$

Also, $$\beta(s)=\frac{1}{4^{s}}\left[\zeta(s,1/4)-\zeta(s,3/4)\right]$$

$$\Gamma(s)\beta(s)=\int_{0}^{\infty}\frac{x^{s-1}e^{-x}}{1+e^{-2x}}dx$$

You know this of course, but I list them to show the relation.

### Re: Integration Contest - Season 5

Sat May 30, 2015 10:02 pm

Posts: 12
Yes I was trying to prove the values of the derivatives of the hurwitz zeta that you've written there. Can you do that? I can't understand what he (adamchik) has done there at all (probably because of my rusty knowledge of bernoulli numbers).

I reduced the integral into something in terms of digamma(1/4) which I can deal with and hurwitz zeta'(1,1/4) which is what I cannot prove from its definition.

### Re: Integration Contest - Season 5

Sat May 30, 2015 10:58 pm
galactus
Global Moderator

Posts: 902
No, I have not tried deriving those values. I just look them up and use them if they arise.

From what I understand, Adamchik used something called Rademcher's formula.

### Re: Integration Contest - Season 5

Sun May 31, 2015 1:45 pm

Posts: 20
Location: Israel
To evaluating $\beta'(1)$ :

Differentiate the functional equation $$\beta(1-s)=\left ( \frac{\pi}{2} \right )^{-s} \sin(\frac{\pi s}{2})\Gamma(s)\beta(s)$$
and put $s=0$ in order to express $\beta'(1)$ in terms of $\beta'(0)$.

Now, we can express the beta function using the Lerch Transcendent $\Phi(z,s,a)=\sum_{n=0}^\infty \frac{z^n}{(a+n)^s}$
as follows: $$\beta(s) = 2^{-s}\Phi(-1,s,\frac{1}{2})$$,
meaning that $$\beta'(s) = -\log(2)\beta(s)+2^{-s} \frac{d}{ds} \Phi(-1,s,\frac{1}{2})$$
and particulary $$\beta'(0) = -\frac{1}{2}\log(2)+\frac{d}{ds} \Phi(-1,s,\frac{1}{2})|_{s=0}$$
The result then follows from $$\frac{\partial}{\partial s}\Phi(-1,0,a) = \log\frac{\Gamma(\frac{a}{2})}{\Gamma(\frac{a+1}{2})\sqrt{2}}$$
which is proven here (page 7, Example 2.6) : http://citeseerx.ist.psu.edu/viewdoc/do ... 1&type=pdf
their proof is based on $\frac{\partial}{\partial s}\Phi(1,0,a) = \log\frac{\Gamma(a)}{\sqrt{2\pi}}$,
proven here (employing known results about Hurwitz's zeta)> (page 26) : http://apps.nrbook.com/bateman/Vol1.pdf

### Re: Integration Contest - Season 5

Sat Jun 06, 2015 11:13 pm
Random Variable Integration Guru

Posts: 381
Problem number something or other

$$\int_{0}^{1} \frac{\cos(1+x^{2})}{1+x^{2}} \, dx = \frac{\pi}{4} - \pi \, \text{S} \left(\sqrt{\frac{2}{\pi}} \right)\text{C} \left(\sqrt{\frac{2}{\pi}} \right)$$

where $\text{S}(z)$ and $\text{C}(z)$ are the Fresnel integrals defined as $$\text{S}(z) = \int_{0}^{z} \sin \left(\frac{\pi x^{2}}{2} \right) \, dx \ , \ \text{C}(z) = \int_{0}^{z} \cos \left(\frac{\pi x^{2}}{2} \right) \, dx$$

That is how Mathematica defines them.

### Re: Integration Contest - Season 5

Tue Jun 09, 2015 11:49 am

Posts: 50
\begin{align}
\int^1_0\frac{\cos(1+x^2)}{1+x^2}{\rm d}x
&=\int^1_0\frac{{\rm d}x}{1+x^2}-\int^1_0\int^1_0\sin(t+tx^2)\ {\rm d}x\ {\rm d}t\\
&=\frac{\pi}{4}-\int^1_0\int^1_0\sin{t}\cos(tx^2)-\cos{t}\sin(tx^2)\ {\rm d}x\ {\rm d}t\\
&=\frac{\pi}{4}-\sqrt{\frac{\pi}{2}}\int^1_0\frac{\sin{t}\ {\rm C}\left(\sqrt{\frac{2t}{\pi}}\right)-\cos{t}\ {\rm S}\left(\sqrt{\frac{2t}{\pi}}\right)}{\sqrt{t}}\ {\rm d}t\\
&=\frac{\pi}{4}-\sqrt{2\pi}\int^1_0\sin(t^2)\ {\rm C}\left(\sqrt{\frac{2}{\pi}}t\right)-\cos(t^2)\ {\rm S}\left(\sqrt{\frac{2}{\pi}}t\right)\ {\rm d}t\\
&=\frac{\pi}{4}-\sqrt{2\pi}\left(\sqrt{\frac{\pi}{2}}{\rm C}\left(\sqrt{\frac{2}{\pi}}\right){\rm S}\left(\sqrt{\frac{2}{\pi}}\right)-\int^1_0\left[\cos(t^2){\rm S}\left(\sqrt{\frac{2}{\pi}}\right)-\cos(t^2){\rm S}\left(\sqrt{\frac{2}{\pi}}\right)\right]\ {\rm d}t\right)\\
&=\frac{\pi}{4}-\pi\ {\rm C}\left(\sqrt{\frac{2}{\pi}}\right){\rm S}\left(\sqrt{\frac{2}{\pi}}\right)
\end{align}
Similarly,
\begin{align}
\int^1_0\frac{\sin(1+x^2)}{1+x^2}\ {\rm d}x
&=\frac{\pi}{2}\left({\rm C}^2\left(\sqrt{\frac{2}{\pi}}\right)-{\rm S}^2\left(\sqrt{\frac{2}{\pi}}\right)\right)
\end{align}

### Re: Integration Contest - Season 5

Tue Jun 09, 2015 3:14 pm
Random Variable Integration Guru

Posts: 381
Nice.

Alternatively,

\begin{align} \left[ \text{C} \left( \sqrt{\frac{2}{\pi}}\right) + i \, \text{S}\left(\sqrt{\frac{2}{\pi}} \right) \right]^{2} &= \frac{2}{\pi} \int_{0}^{1} \int_{0}^{1} e^{i(x^{2}+t^{2})} \ dx \ dt \\ &= \frac{4}{\pi} \int_{0}^{1} \int_{0}^{t} e^{i(x^{2}+t^{2})} \, dx \, dt \\ &= \frac{4}{\pi} \int_{0}^{\pi /4} \int_{0}^{\sec \theta} e^{i r^{2}} r \, dr \, d \theta \\ &= \frac{2 i}{\pi} \int_{0}^{\pi /4} \left(1- e^{i \sec^{2} \theta} \right) \, d \theta \\ &= \frac{2i}{\pi} \left(\frac{\pi}{4} - \int_{0}^{1} \frac{e^{i(1+u^{2})}}{1+u^{2}} \ du \right) \end{align}

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