1. Originally Posted by chiph588@
Problem $\displaystyle (\star\star\star\star)$: For $\displaystyle |x|<1$, compute

$\displaystyle \displaystyle \sum_{s=2}^{\infty}\frac{x^s\zeta(s)}{s}$
Solution:

EDIT: Already beaten to the answer by simplependulum, but a different method.

Spoiler:

We begin by noticing that for $\displaystyle x\notin -\mathbb{N}$ we have that

$\displaystyle \displaystyle \Gamma(-x)=\frac{e^{\gamma x}}{-x}\prod_{n=1}^{\infty}\left(1-\frac{x}{n}\right)^{-1}e^{\frac{-x}{n}}$

(I prove this here). Taking the logarithm of both sides gives

$\displaystyle \displaystyle \log\left(\Gamma(-x)\right)=\gamma x-\log(-x)+\sum_{n=1}^{\infty}\left(-\log\left(1-\frac{x}{n}\right)-\frac{x}{n}\right)$

Or, said differently

$\displaystyle \displaystyle \log\left(\Gamma(1-x)\right)-\gamma x=\sum_{n=1}^{\infty}\left(-\log\left(1-\frac{x}{n}\right)-\frac{x}{n}\right)\quad\mathbf{(1)}$

But, note that since $\displaystyle |x|<1$ we have that

\displaystyle \displaystyle \begin{aligned}\sum_{n=1}^{\infty}\left(-\log\left(1-\frac{x}{n}\right)-\frac{x}{n}\right) &=\sum_{n=1}^{\infty}\left(\sum_{m=0}^{\infty}\fra c{x^{m+1}}{n^{m+1} (m+1)}-\frac{x}{n}\right)\\ &= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{x^{m+1 }}{n^{m+1}(m+1)}\end{aligned}

Note though that for each fixed $\displaystyle n$ the inner sum converges absolutely since $\displaystyle |x|<1$. Thus, we may rearrange the series (see Apostol pg. 202) to get

\displaystyle \displaystyle \begin{aligned}\sum_{n=1}^{\infty}\sum_{m=1}^{\inf ty}\frac{x^{m+1}}{n^{m+1}(m+1)} &= \sum_{m=1}^{\infty}\frac{x^{m+1}}{m+1}\sum_{n=1}^{ \infty}\frac{1}{n^{m+1}}\\ &= \sum_{m=1}^{\infty}\frac{x^{m+1}\zeta(m+1)}{m+1}\\ &= \sum_{m=2}^{\infty}\frac{x^m \zeta(m)}{m}\end{aligned}

Plugging this into $\displaystyle \mathbf{(1)}$ gives then that

$\displaystyle \displaystyle \sum_{m=2}^{\infty}\frac{x^m\zeta(m)}{m}=\log\left (\Gamma(1-x)\right)-\gamma x$

for all $\displaystyle |x|<1$.

2. Originally Posted by Drexel28
Solution:

EDIT: Already beaten to the answer by simplependulum, but a different method.

Spoiler:

We begin by noticing that for $\displaystyle x\notin -\mathbb{N}$ we have that

$\displaystyle \displaystyle \ln\left(\Gamma(-x)\right)=\frac{e^{\gamma x}}{-x}\prod_{n=1}^{\infty}\left(1-\frac{x}{n}\right)^{-1}e^{\frac{-x}{n}}$

(I prove this here). Taking the logarithm of both sides gives

$\displaystyle \displaystyle \log\left(\Gamma(-x)\right)=\gamma x-\log(-x)+\sum_{n=1}^{\infty}\left(-\ln\left(1-\frac{x}{n}\right)-\frac{x}{n}\right)$

Or, said differently

$\displaystyle \displaystyle \log\left(\Gamma(1-x)\right)-\gamma x=\sum_{n=1}^{\infty}\left(-\ln\left(1-\frac{x}{n}\right)-\frac{x}{n}\right)\quad\mathbf{(1)}$

But, note that since $\displaystyle |x|<1$ we have that

\displaystyle \displaystyle \begin{aligned}\sum_{n=1}^{\infty}\left(-\ln\left(1-\frac{x}{n}\right)-\frac{x}{n}\right) &=\sum_{n=1}^{\infty}\left(\sum_{m=0}^{\infty}\fra c{x^{m+1}}{n^{m+1} (m+1)}-\frac{x}{n}\right)\\ &= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{x^{m+1 }}{n^{m+1}(m+1)}\end{aligned}

Note though that for each fixed $\displaystyle n$ the inner sum converges absolutely since $\displaystyle |x|<1$. Thus, we may rearrange the series (see Apostol pg. 202) to get

\displaystyle \displaystyle \begin{aligned}\sum_{n=1}^{\infty}\sum_{m=1}^{\inf ty}\frac{x^{m+1}}{n^{m+1}(m+1)} &= \sum_{m=1}^{\infty}\frac{x^{m+1}}{m+1}\sum_{n=1}^{ \infty}\frac{1}{n^{m+1}}\\ &= \sum_{m=1}^{\infty}\frac{x^{m+1}\zeta(m+1)}{m+1}\\ &= \sum_{m=2}^{\infty}\frac{x^m \zeta(m)}{m}\end{aligned}

Plugging this into $\displaystyle \mathbf{(1)}$ gives then that

$\displaystyle \displaystyle \sum_{m=2}^{\infty}\frac{x^m\zeta(m)}{m}=\log\left (\Gamma(1-x)\right)-\gamma x$

for all $\displaystyle |x|<1$.

That's how I did it!

Originally Posted by simplependulum
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 ...
I've seen a short but messy proof of this fact. I shall spare everyone the headache of reading my solution!

3. Here's an easy one to keep the thread going. Evaluate

$\displaystyle \int_0^\infty \frac{\log x}{1+x^2} dx$

4. Originally Posted by Bruno J.
Here's an easy one to keep the thread going. Evaluate

$\displaystyle \int_0^\infty \frac{\log x}{1+x^2} dx$
Easy way (more general):
Spoiler:

Let

$\displaystyle \displaystyle I(a)=\int_0^{\infty}\frac{\log(x)}{x^2+a^2}\text{ }a>0$

Let $\displaystyle x=\frac{a}{t}$ to get

\displaystyle \displaystyle \begin{aligned}I(a) &= \frac{\log(a)}{a}\int_0^{\infty}\frac{dt}{1+t^2}-\frac{1}{a}\int_0^{\infty}\frac{\log(t}{1+t^2}\tex t{ }dt\end{aligned}

But, let $\displaystyle \displaystyle y=\frac{1}{u}$ in this second integral to get that

$\displaystyle \displaystyle \int_0^{\infty}\frac{\log(t)}{1+t^2}=-\int_0^{\infty}\frac{\log(u)}{1+u^2}\text{ }du$

So that

$\displaystyle \displaystyle \int_0^{\infty}\frac{\log(t)}{1+t^2}\text{ }dt=0$

Thus,

$\displaystyle \displaystyle I(a)=\frac{\pi\log(a)}{2a}$

To spice things up:

Spoiler:

Consider

$\displaystyle \displaystyle f(z)=\frac{\log(z)}{z^2+a^2}$

and for $\displaystyle 0<\varepsilon<R$ define the contour $\displaystyle \Omega_{\varepsilon,R}$ to be

$\displaystyle \displaystyle \Omega_{\epsilon,R}=\Gamma_R\cup\gamma_{\varepsilo n}\cup[-R,-\varepsilon]\cup[\varepsilon,R]$

Where $\displaystyle \Gamma_R$ is the half-annulus $\displaystyle \Gamma_R=\varepsilon<|z|<R$ and $\displaystyle \gamma_{\varepsilon}$ is the semi-circle of radius $\displaystyle \varepsilon$ centered at $\displaystyle 0$. It's evident that

$\displaystyle \displaystyle \left|\int_{\gamma_{\varepsilon}}f(z)\text{ }dz\right|\to0\quad \varepsilon\to 0$

and

$\displaystyle \displaystyle \left|\int_{\Gamma_R}f(z)\text{ }dz\right|\leqslant \frac{|\log(R)|+\pi}{R^2-1}\pi R\to 0\quad R\to\infinity$

But, by the residue theorem

$\displaystyle \displaystyle \int_{\Omega_{R,\varepsilon}}f(z)\text{ }dz=2\pi i \text{Res}\left(f;ia\right)=\frac{\pi}{a}\left(\lo g(a)+\frac{\pi i}{2}\right)$

And thus it follows that

$\displaystyle \displaystyle \int_{-\infty}^{\infty}\frac{\log(z)}{z^2+a^2}\text{ }dz=\frac{\pi}{a}\left(\log(a)+\frac{\pi i}{2}\right)$

But,

\displaystyle \displaystyle \begin{aligned}\int_{-\infty}^{0}f(z)\text{ }dz &= \int_{-\infty}^{0}\frac{\log(-x)+\pi i}{x^2+a^2}\text{ }dx\\ &= \int_0^{\infty}\frac{\log(x)}{x^2+a^2}\text{ }dx+\frac{\pi^2 i}{2a}\end{aligned}

From where it evidently follows that

$\displaystyle \displaystyle \int_0^{\infty}\frac{\log(x)}{x^2+a^2}\text{ }dx=\frac{\pi\log(a)}{2a}$

5. Good! I prefer your first method, since it works fine and it's much easier.

Now how about $\displaystyle \int_0^1 \frac{\log(1+x)}{1+x^2}$?

By the way, you might want to be more precise in the way you've defined your contour. As it is, $\displaystyle \Gamma_R$ is a domain and not a contour. You also didn't give an orientation.

6. let $\displaystyle x = \frac{1-t}{1+t}$

EDIT: or let $\displaystyle I(a) = \int_{0}^{1} \frac{\ln(ax+1)}{1+x^{2}} \ dx$ and differentiate

7. Originally Posted by Random Variable
let $\displaystyle x = \frac{1-t}{1+t}$
Where did you come up with this substitution?

8. Originally Posted by Bruno J.
Good! I prefer your first method, since it works fine and it's much easier.

Now how about $\displaystyle \int_0^1 \frac{\log(1+x)}{1+x^2}$?
EDIT: Beat by Random Variable
The classic way to do this is the substitution $\displaystyle x=\frac{1-t}{1+t}$ to get, if $\displaystyle I$ equals our integral, that $\displaystyle \displaystyle I=\int_0^1\frac{\log(2)}{t^2+1}-I$. So that $\displaystyle \displaystyle I=\frac{\log(2)\pi}{8}$. Now, solve

$\displaystyle \displaystyle \int_0^{\infty}\frac{\log(1+x)}{1+x^2}\text{ }dx$

9. Originally Posted by Drexel28
EDIT: Beat by Random Variable
The classic way to do this is the substitution $\displaystyle x=\frac{1-t}{1+t}$ to get, if $\displaystyle I$ equals our integral, that $\displaystyle \displaystyle I=\int_0^1\frac{\log(2)}{t^2+1}-I$. So that $\displaystyle \displaystyle I=\frac{\log(2)\pi}{8}$. Now, solve

$\displaystyle \displaystyle \int_0^{\infty}\frac{\log(1+x)}{1+x^2}\text{ }dx$
Sorry. I should only post complete solutions.

I really can't explain why that substitution works. Can you?

10. Originally Posted by Random Variable
I really can't explain why that substitution works. Can you?
Then what made you think of it?

11. Originally Posted by Random Variable
Sorry. I should only post complete solutions.

I really can't explain why that substitution works. Can you?
I can, what's not to understand. Letting $\displaystyle \displaystyle x=\frac{1-t}{1+t}$ we get (with all the intermediary steps) that

\displaystyle \displaystyle \begin{aligned}\int_0^1\frac{\log(1+x)}{1+x^2}\tex t{ }dx &= \int_1^0\frac{\log\left(1+\frac{1-t}{1+t}\right)}{1+\frac{(1-t)^2}{(1+t)^2}}\frac{-2}{(1+t)^2}\text{ }dt\\ &= \int_0^1\frac{2\log\left(\frac{2}{x+1}\right)}{(1+ t)^2+(1-t)^2}\\ &= \int_0^1\frac{\log(2)-\log(1+t)}{1+t^2}\text{ }dt\end{aligned}

12. Originally Posted by Drexel28
$\displaystyle \displaystyle \int_0^{\infty}\frac{\log(1+x)}{1+x^2}\text{ }dx$
Spoiler:

\displaystyle \begin{aligned}\displaystyle \int_0^{\infty}\frac{\log(1+x)}{1+x^2} dx &= \int_0^{1}\frac{\log(1+x)}{1+x^2} dx+\int_1^{\infty}\frac{\log(1+x)}{1+x^2} dx\\ &= \frac18\pi\log2 + \int_0^1 \frac{\log(1+1/x)}{1+x^2} dx \quad \text{ (here we let } x \mapsto 1/x \text{)}\\ &= \frac18\pi\log2 + \int_0^{1}\frac{\log(1+x)}{1+x^2} dx - \int_0^{1}\frac{\log(x)}{1+x^2} dx\\ &= \frac14\pi\log2 + G \end{aligned}

Where $\displaystyle G$ is Catalan's constant. See here for why.

13. No. What I mean is what about the integral would suggest that substitution.

14. It's a classic Putnam integral. That's the only reason I knew.

15. Originally Posted by chiph588@
Spoiler:

\displaystyle \begin{aligned}\displaystyle \int_0^{\infty}\frac{\log(1+x)}{1+x^2} dx &= \int_0^{1}\frac{\log(1+x)}{1+x^2} dx+\int_1^{\infty}\frac{\log(1+x)}{1+x^2} dx\\ &= \frac18\pi\log2 + \int_0^1 \frac{\log(1+1/x)}{1+x^2} dx \text{(here we let } x \mapsto 1/x \text{)}\\ &= \frac18\pi\log2 + \int_0^{1}\frac{\log(1+x)}{1+x^2} dx - \int_0^{1}\frac{\log(x)}{1+x^2} dx\\ &= \frac14\pi\log2 + G \end{aligned}

Where $\displaystyle G$ is Catalan's constant. See here for why.

Right, that's what I was thinking.

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