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## Series Contest (SC4)

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### Series Contest (SC4)

Thu May 15, 2014 6:10 am

Posts: 850
Location: Jaipur, India

Welcome! I announce the Beginning of another Series Contest (SC4)! I am not going to post rules again and again as most of our members are familiar with the idea of contests. For newcomers I would suggest to go through the previous contests before taking part.

Remember numbering your problems while posting. Have fun and enjoy!

### Re: Series Contest (SC4)

Thu May 15, 2014 6:12 am

Posts: 850
Location: Jaipur, India

Problem 1

Show that $\displaystyle \sum_{k=1}^\infty \text{Ci}(2\pi k n )=\frac{1}{2}\left\{\psi_0(n)-\log(n)+\frac{1}{2n} \right\}$

where $\displaystyle \text{Ci}(x)=-\int_{x}^{\infty}\frac{\cos(t)}{t}dt$

### Re: Series Contest (SC4)

Fri May 16, 2014 8:56 am
Random Variable Integration Guru

Posts: 381
$\displaystyle \sum_{k=1}^{\infty} \text{Ci}(2 \pi k n) = - \sum_{k=1}^{\infty} \int_{2 \pi k n}^{\infty} \frac{\cos t}{t} \ dt$

$\displaystyle = - \sum_{k=1}^{\infty} \int_{2 \pi kn}^{\infty} \int_{0}^{\infty} \cos t e^{-tx} \ dx \ dt = - \sum_{k=0}^{\infty} \int_{0}^{\infty} \int_{2 \pi k n}^{\infty} \cos t e^{-xt} dt \ dx$

$\displaystyle = -\sum_{k=1}^{\infty} \int_{0}^{\infty} e^{-2 \pi k n x} \frac{x}{1+x^{2}} \ dx$

$= \displaystyle -\int_{0}^{\infty} \frac{x}{1+x^{2}} \sum_{k=1}^{\infty} e^{-2 \pi k n x} \ dx = -\int_{0}^{\infty} \frac{x}{1+x^{2}} \frac{e^{- 2 \pi n}}{1-e^{-2 \pi n}} \ dx$

$\displaystyle = -\int_{0}^{\infty} \frac{x}{1+x^{2}} \frac{1}{e^{2 \pi nx}-1} \ dx$

Without the $n$, all we would need to do is differentiate Binet's log gamma formula.

If someone would like to continue from here, go ahead. Otherwise I'll try and finish it later.

### Re: Series Contest (SC4)

Fri May 16, 2014 7:06 pm
Random Variable Integration Guru

Posts: 381
I'm stupid. All you need to do is make a substitution.

$\displaystyle \sum_{k=1}^{\infty} \text{Ci}(2 \pi k n) = -\int_{0}^{\infty} \frac{x}{1+x^{2}} \frac{1}{e^{2 \pi nx}-1} \ dx$

$\displaystyle = -\int_{0}^{\infty} \frac{u}{n^{2}+u^{2}} \frac{1}{e^{2 \pi u}-1} \ du$

And differentiating Binet's formula,

$\displaystyle \psi(z) = \ln z + \left(z- \frac{1}{2} \right) \frac{1}{z} - 1 - 2 \int_{0}^{\infty} \frac{t}{z^{2}+t^{2}} \frac{1}{e^{2 \pi t}-1} \ dt$

Therefore,

$\displaystyle \sum_{k=1}^{\infty} \text{Ci}(2 \pi k n) = \frac{1}{2} \left( \psi(n) - \ln(n) + \frac{1}{2n} \right)$

### Re: Series Contest (SC4)

Fri May 16, 2014 7:59 pm
galactus
Global Moderator

Posts: 902
Very good G man

That's a nice way to go about it.

Here is a formula that may come in handy here or there:

it appears to be a Binet type formula

$$\psi(a)=log(a)-\frac{1}{2a}+2\sum_{n=1}^{\infty}\left[\cos(2\pi na)Ci(2\pi na)+\sin(2\pi na)si(2\pi na)\right]...(1)$$

Let $a=n$ and get:

$$\psi(n)=log(n)-\frac{1}{2n}+2\sum_{m=1}^{\infty}Ci(2n\pi m)$$

(1) comes about by using the Binet formula $$log\Gamma(u)=(u-1/2)log(u)-u+1/2log(2\pi)+\frac{1}{\pi}\sum_{n=1}^{\infty}\int_{0}^{\infty}\frac{\sin(2\pi nx)}{n(x+u)}dx$$

and diffing w.r.t u.

an interesting side note is:

$$2\int_{0}^{\infty}\frac{\tan^{-1}(x/u)}{e^{2\pi x}-1}dx=\frac{1}{\pi}\sum_{n=1}^{\infty}\int_{0}^{\infty}\frac{\sin(2\pi nx)}{n(x+u)}dx$$

The Abel-Plana sum formula can be a handy thing(I used it for that other problem S posted).

By using $f(n)=\frac{1}{n+u)^{s}}$, then:

$$\zeta(s,u)=\sum_{n=0}^{\infty}\frac{1}{(n+u)^{s}}=\frac{1}{2u^{s}}+\frac{u^{1-s}}{s-1}+i\int_{0}^{\infty}\frac{(u+ix)^{-s}-(u-ix)^{-s}}{e^{2\pi x}-1}dx$$

But, $$(u+ix)^{-s}-(u-ix)^{-s}=r^{-s}(e^{-is\theta}-e^{is\theta})=\frac{2}{i(u^{2}+x^{2})^{s/2}}\sin(s\cdot \tan^{-1}(x/u))$$

diffing and so forth brings one around to the Binet thing. I am sure you fellas already know this, but it is a cool thing to keep in the toolbelt.

### Re: Series Contest (SC4)

Sat May 17, 2014 2:05 am
Random Variable Integration Guru

Posts: 381
Problem 2

$\displaystyle \sum_{k=1}^{\infty} \frac{(2k-1)^{4n+1}}{1+e^{(2k-1) \pi}} = \frac{1}{2} \frac{1}{\pi^{4n+2}} \left(1-\frac{2}{2^{4n+2}}\right) (4n+1)! \zeta(4n+2) = \frac{2^{4n+1}-1}{8n+4} B_{4n+2}$

### Re: Series Contest (SC4)

Sat May 17, 2014 9:12 am
zaidalyafey
Global Moderator

Posts: 354
Random Variable wrote:
Problem 2

$\displaystyle \sum_{n=1}^{\infty} \frac{(2k-1)^{4n+1}}{1+e^{(2k-1) \pi}} = \frac{1}{2} \frac{1}{\pi^{4n+2}} \left(1-\frac{2}{2^{4n+2}}\right) (4n+1)! \zeta(4n+2) = \frac{2^{4n+1}-1}{8n+4} B_{4n+2}$

so the index of summation must be $k=1$ , right ?
$\displaystyle \sum_{k\geq 0}\frac{f^{(k)} (s)}{(s+1)_k} (-s)^{k}= \int^{1}_0x^{s} f(s x) \left( \frac{x}{s}\right) \, dx$

Wanna learn what we discuss , see tutorials

### Re: Series Contest (SC4)

Sat May 17, 2014 11:20 am
Random Variable Integration Guru

Posts: 381
zaidalyafey wrote:
Random Variable wrote:
Problem 2

$\displaystyle \sum_{n=1}^{\infty} \frac{(2k-1)^{4n+1}}{1+e^{(2k-1) \pi}} = \frac{1}{2} \frac{1}{\pi^{4n+2}} \left(1-\frac{2}{2^{4n+2}}\right) (4n+1)! \zeta(4n+2) = \frac{2^{4n+1}-1}{8n+4} B_{4n+2}$

so the index of summation must be $k=1$ , right ?

Yes.

### Re: Series Contest (SC4)

Sat May 17, 2014 12:31 pm

Posts: 32
the sum can be gotten from this eq:
$$\sum\limits_{n{\rm{ = }}0}^\infty {\frac{{\left( {n + \frac{1}{2}} \right)}}{{{{\left( {n + \frac{1}{2}} \right)}^4} + {a^4}}}} \frac{1}{{{e^{2\pi n + 1}} + 1}} = - \frac{\pi }{{2{a^2}}} \times \frac{1}{{\exp \left( {2\sqrt 2 a\pi } \right) + 2\exp \left( {\sqrt 2 a\pi } \right)\cos \left( {\sqrt 2 a\pi } \right) + 1}} + \frac{\pi }{{8{a^2}}} \\ \, \, \, - \frac{1}{{2{a^2}}}{\mathop{\rm Im}\nolimits} \psi \left( {\frac{1}{2} + \frac{a}{{\sqrt 2 }} + \frac{a}{{\sqrt 2 }}i} \right)$$
Generalized Abel-Plana Formula will work on it (reference :The generalized Abel-Plana formula with applications to Bessel functions and Casimir effect,Ramanujan'lost book),but Yelusalen gave another proof by using Complex Analysis,he considered the contour integral
$$\int_C {\frac{z}{{{z^4} + {a^4}}}\frac{1}{{{e^{2\pi z}} + 1}}} \frac{1}{{{e^{ - 2\pi iz}} + 1}}dz$$,the contour of which is $$[0,R] + {C_R} + [Ri,0],R \to \infty$$,bypassing all poles on the borders.
Last edited by zaidalyafey on Sat May 17, 2014 1:29 pm, edited 1 time in total.
Reason: Code is too long to be included in one line

### Re: Series Contest (SC4)

Mon May 19, 2014 11:22 am
Random Variable Integration Guru

Posts: 381
Marko Riedel evaluated the second series using the inverse Mellin transform.

http://math.stackexchange.com/questions ... 188#740188

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