1. Originally Posted by chiph588@
I did the same thing except got $\displaystyle I(\alpha,\beta) = 2\arctanh^2\left(\frac{a-1}{\sqrt{a^2-1}}\right)-2\arctanh^2\left(\frac{b-1}{\sqrt{b^2-1}}\right)$.
I get that our integral is equal to $\arccos(\beta)^2-\arccos(\alpha)^2$.

2. $2 \int^{\pi /2}_{0} \sec(x) \ln \big(\frac{1+ \alpha \cos x}{1+\beta \cos x}\big) \ dx = 2 \int_{0}^{\pi /2} \int^{\alpha}_{\beta} \frac{dt \ dx }{1+t \cos x }}$

3. Originally Posted by Drexel28
Yes, the obvious way is to use CA. One question, if you know CA how do you not know multiple integrals?
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.

Originally Posted by Drexel28
Woah, your answer looks completely different than mine. Are you sure it's correct? Not doubting your, but it's just really different.
I checked one value numerically with wolfram|alpha (for $\alpha = 0.5\:\:\:\:,\:\:\:\:\beta=0$. It can't do the integral in closed form ) 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.

Originally Posted by chiph588@
I did magic differentiation and got $\displaystyle I(\alpha,\beta) = 4\left(\tanh^{-1}\left(\frac{a-1}{\sqrt{a^2-1}}\right)\right)^2-4\left(\tanh^{-1}\left(\frac{b-1}{\sqrt{b^2-1}}\right)\right)^2$.

Edit: Oh, that's even better:

Originally Posted by Drexel28
I get that our integral is equal to $\arccos(\beta)^2-\arccos(\alpha)^2$.

4. Originally Posted by Drexel28
I get that our integral is equal to $\arccos(\beta)^2-\arccos(\alpha)^2$.
Hmm, it turns out $\displaystyle 4\left(\tanh^{-1}\left(\frac{a-1}{\sqrt{a^2-1}}\right)\right)^2 = -\arccos^2(a)$, for $a>0$. I'd be interested to see what you did.

5. Originally Posted by Random Variable
$2 \int^{\pi /2}_{0} \sec(x) \ln \big(\frac{1+ \alpha \cos x}{1+\beta \cos x}\big) \ dx = 2 \int_{0}^{\pi /2} \int^{\alpha}_{\beta} \frac{dt \ dx }{1+t \cos x }}$

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

\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}

6. Originally Posted by chiph588@
Hmm, it turns out $\displaystyle 4\left(\tanh^{-1}\left(\frac{a-1}{\sqrt{a^2-1}}\right)\right)^2 = -\arccos^2(a)$, for $a>0$. I'd be interested to see what you did.
Well, it's commonly known (I should have noticed this in Unbeatable's answer) that $\displaystyle \arccos(x)=\frac{1}{2}\arctan\left(\frac{\sqrt{1-x}}{\sqrt{1+x}}\right)$. Maybe there's a way to use this to derive your result. Using this one can see that Unbeatable's answer is totally correct.

7. Originally Posted by Drexel28
Well, it's commonly known (I should have noticed this in Unbeatable's answer) that $\displaystyle \arccos(x)=\frac{1}{2}\arctan\left(\frac{\sqrt{1-x}}{\sqrt{1+x}}\right)$. Maybe there's a way to use this to derive your result. Using this one can see that Unbeatable's answer is totally correct.
The $\frac{1}{2}$ should have been on the other side of the equality.

It follows that:

$\tanh^{-1}\frac{a-1}{\sqrt{a^2-1}} = -i\tan^{-1}\sqrt\frac{1-a}{1+a} = -\frac{1}{2}i\arccos(a)$, deriving Chiph's result

8. Is the answer still valid if $\alpha, \beta >1$?

Originally Posted by Drexel28
Yes, this is how I did it. And then interchanging th order of integration we get

\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}

9. Originally Posted by Random Variable
Is the answer still valid if $\alpha, \beta >1$?
if $a>1$ you can use $\arccos(a) = -i \ln(a+\sqrt{a^2-1})$

10. But the integrand is real-valued when $\alpha, \beta > 0$. That doesn't make any sense, does it?

11. Originally Posted by Random Variable
But the integrand is real-valued when $\alpha, \beta > 0$. That doesn't make any sense, does it?
But you're squaring, and thus the $i$ disappears

12. Oh, I thought a complex number was being squared, not necessarily a purely imaginary number. My bad.

13. Originally Posted by Drexel28
Problem $\mathbf{(\star\star)}$: Compute

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2+a^2}$
Spoiler:

We have $\displaystyle \frac{\sin\pi a}{\pi a} = \prod_{n=1}^\infty \left(1-\frac{a^2}{n^2}\right) \implies \frac{\sinh \pi a}{\pi a} = \frac{\sin\pi ia}{\pi ia} = \prod_{n=1}^\infty \left(1+\frac{a^2}{n^2}\right)$.

Taking logs and differentiating, we get $\displaystyle \left(\log\frac{\sinh \pi a}{\pi a}\right)' = \sum_{n=1}^\infty \left(\log\left\{1+\frac{a^2}{n^2}\right\}\right)' = \sum_{n=1}^\infty \frac{2a}{n^2+a^2}$.

Thus $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2+a^2} = \frac1{2a}\left[\pi\coth\pi a-\frac1a\right]$.

14. Problem $(\star\star\star\star)$: For $|x|<1$, compute

$\displaystyle \sum_{s=2}^\infty \frac{\zeta(s)}sx^s$

15. The answer is $\displaystyle \ln[ \Gamma(1-x) ] - \gamma x$ 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 :

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

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

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

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

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

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

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

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

Let's get back to your problem :

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

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

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

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

Integrating it to obtain the result :

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

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

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

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

Page 5 of 8 First 12345678 Last