I get that our integral is equal to $\displaystyle \arccos(\beta)^2-\arccos(\alpha)^2$.

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- Dec 27th 2010, 12:32 PMDrexel28
- Dec 27th 2010, 12:34 PMRandom Variable
$\displaystyle 2 \int^{\pi /2}_{0} \sec(x) \ln \big(\frac{1+ \alpha \cos x}{1+\beta \cos x}\big) \ dx = 2 \int_{0}^{\pi /2} \int^{\alpha}_{\beta} \frac{dt \ dx }{1+t \cos x }} $

- Dec 27th 2010, 12:38 PMUnbeatable0
We don't use multiple integrals in complex analysis (at least in the course in the university in which I study), and so it happened that I'll learn multiple integrals only after complex analysis.

I checked one value numerically with wolfram|alpha (for $\displaystyle \alpha = 0.5\:\:\:\:,\:\:\:\:\beta=0$. It can't do the integral in closed form (Surprised)) and it seemed correct. Probably you can differentiate your answer w.r.t. alpha, my answer (although differentiating my answer looks like hell) and see they're the same.

Your form looks much nicer (Rofl)

**Edit**: Oh, that's even better:

(Bigsmile) - Dec 27th 2010, 12:40 PMchiph588@
- Dec 27th 2010, 12:43 PMDrexel28

Yes, this is how I did it. And then interchanging th order of integration we get

$\displaystyle \displaystyle \begin{aligned}2\int_{\beta}^{\alpha}\int_0^{\frac {\pi}{2}}\frac{1}{1+t\cos(x)}\text{ }dx\text{ }dt &= 2\int_{\beta}^{\alpha}\int_0^1\frac{2}{1+t\frac{1-u^2}{1+u^2}}\frac{2}{1+u^2}\text{ }du\text{ }dx\\ &= 4\int_{\beta}^{\alpha}\int_0^1\frac{1}{1-t}\frac{1}{\frac{1+t}{1-t}+u^2}\text{ }du\text{ }dt\\ &= 4\int_{\beta}^{\alpha}\arctan\left(\sqrt{\frac{1-t}{1+t}\right)\frac{1}{\sqrt{1-t^2}}\text{ }dt\\ &= 2\int_{\beta}^{\alpha}\frac{\arccos(w)}{\sqrt{1-w^2}}\text{ }dw\\ &= \arccos(\beta)^2-\arccos(\alpha)^2\end{aligned}$ - Dec 27th 2010, 12:56 PMDrexel28
Well, it's commonly known (I should have noticed this in

**Unbeatable**'s answer) that $\displaystyle \displaystyle \arccos(x)=\frac{1}{2}\arctan\left(\frac{\sqrt{1-x}}{\sqrt{1+x}}\right)$. Maybe there's a way to use this to derive your result. Using this one can see that**Unbeatable**'s answer is totally correct. - Dec 27th 2010, 01:03 PMUnbeatable0
- Dec 27th 2010, 01:37 PMRandom Variable
- Dec 27th 2010, 01:40 PMUnbeatable0
- Dec 27th 2010, 02:08 PMRandom Variable
But the integrand is real-valued when $\displaystyle \alpha, \beta > 0$. That doesn't make any sense, does it?

- Dec 27th 2010, 02:11 PMUnbeatable0
- Dec 27th 2010, 02:18 PMRandom Variable
Oh, I thought a complex number was being squared, not necessarily a purely imaginary number. My bad.

- Dec 27th 2010, 02:41 PMchiph588@
- Dec 27th 2010, 04:40 PMchiph588@
**Problem $\displaystyle (\star\star\star\star)$:**For $\displaystyle |x|<1 $, compute

$\displaystyle \displaystyle \sum_{s=2}^\infty \frac{\zeta(s)}sx^s $ - Dec 27th 2010, 07:10 PMsimplependulum
The answer is $\displaystyle \displaystyle \ln[ \Gamma(1-x) ] - \gamma x$ But the way I evaluate is to use the infinite product of Gamma function which makes me confused .

For this reason , i attempt to solve it without using the infinite product :

First , I need to obtain the expression of Digamma function :

$\displaystyle \displaystyle \Gamma'(z) = \int_0^{\infty} e^{-t} t^{z-1} \ln(t) ~dt $

Since $\displaystyle \displaystyle \ln(t) = \int_0^{\infty} \frac{e^{-x} - e^{-xt} }{x}~dx $ substituting it we obtain :

$\displaystyle \displaystyle \Gamma'(z) = \int_0^{\infty}\int_0^{\infty}(e^{-x} - e^{-xt})e^{-t} t^{z-1} ~dt~\frac{dx}{x} $

$\displaystyle \displaystyle = \Gamma(z) \int_0^{\infty} ( e^{-x} - \frac{1}{(x+1)^z} ) ~\frac{dx}{x} $

Then $\displaystyle \displaystyle \frac{ \Gamma'(1)}{\Gamma(1)} - \frac{ \Gamma'(z)}{\Gamma(z)} = \int_0^{\infty} \left( \frac{1}{(x+1)^z} - \frac{1}{x+1} \right) ~ \frac{dx}{x} $

Sub. $\displaystyle x+1 = e^{y}$ we have

$\displaystyle \displaystyle \frac{ \Gamma'(1)}{\Gamma(1)} - \frac{ \Gamma'(z)}{\Gamma(z)} = \int_0^{\infty} \frac{e^{(1-z)y} - 1 }{ e^y -1 }~dy $

Therefore , $\displaystyle \displaystyle \int_0^{\infty} \frac{e^{(1-z)y} - 1 }{ e^y -1 }~dy = -\gamma - \frac{ \Gamma'(z)}{\Gamma(z)} $

Let's get back to your problem :

Let $\displaystyle \displaystyle F(x) = \sum_{s=2}^{\infty} \zeta(s) x^{s-1}$

$\displaystyle \displaystyle = \int_0^{\infty} \sum_{s=2}^{\infty} \frac{(xt)^{s-1}}{(s-1)!} ~\frac{dt}{e^t - 1 } $

$\displaystyle \displaystyle = \int_0^{\infty} \frac{ e^{xt} - 1 }{e^t - 1 } ~dt $

$\displaystyle \displaystyle = - \gamma - \frac{ \Gamma'(1-x)}{\Gamma(1-x)} $

Integrating it to obtain the result :

$\displaystyle \displaystyle \sum_{s=2}^{\infty} \frac{\zeta(s)}{s} x^s = \ln[\Gamma(1-x)] - \gamma x $

Using the similar method , Prove this formula ( Gauss's Multiplication Formula ) :

$\displaystyle \displaystyle \Gamma(nz) = n^{nz - \frac{1}{2} } (2\pi)^{\frac{1-n}{2} \prod_{k=0}^{n-1} \Gamma{\left( z + \frac{k}{n} \right)} ~ n \in \mathbb{N} $

ps I don't know how many stars it should be given ...