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@Frpzzd :P

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@Frpzzd By the way, your newest question is most nearly a dupe ;)

Thank you!

:P

Btw, have you seen this new user called Nilknarf? @Frpzzd

@ξξξ Hello and welcome to my realm!

@Frpzzd By the way, I think the best telescoping sums are the ones involving arctan of a rational quadratic and the fundamental theorem of calculus.

Since you know of the geometric series, I think you'll like the following integral @Frpzzd whenever you decide to read it.

$$\begin{align}\int_0^\infty\frac{x^s}{e^x-1}\ dx& =\int_0^\infty x^se^{-x}\left(\frac1{1-e^{-x}}\right)\ dx\\& =\int_0^\infty x^se^{-x}\sum_{k=0}^\infty e^{-kx}\ dx\\& =\sum_{k=1}^\infty\int_0^\infty x^se^{-kx}\ dx\\& =\sum_{k=1}^\infty\frac1{k^{s+1}}\int_0^\infty u^se^{-u}\ du\\& =\zeta(s+1)\Gamma(s+1)\end{align}$$

Where we use the u-substitution $u=kx$

@SBM May also find this integral interesting

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@SimplyBeautifulArt Zeta and gamma together; interesting

:-)

how do you evaluate $$\int \frac{1}{\sqrt{1-e^{2x}}} ~ \mathrm{d} x$$ ?

@SBM Probably can't

not without bounds

is it something like a hyperbolic function?

idk

should we try substitution?

$$\int\frac1{\sqrt{1-e^{2x}}}~\mathrm dx= \int\frac{e^{-x}}{\sqrt{e^{-2x}-1}}~\mathrm dx =\int\frac{-1}{\sqrt{u^2-1}}~\mathrm du$$

The last one is doable with trig substitution

So if I assume $u = \sin k \implies \mathrm du = \cos k ~\mathrm dk$

now that'd make it like $\int - \mathrm d k$

Hm... I don't think so

oh

Use hyperbolic trig functions

ok so I guess \[ \int \frac{1}{\sqrt{u^2 - 1}} ~ \mathrm d u = \operatorname{arcosh} u + C \] ?

is it?

then I'd probably get $-\operatorname{arcosh} u + C$

I think that's right

which'd be like $\ln |e^{-2x} + \sqrt{e^{-2x} - 1}| + C$

yeah

Hm...

Nope, nvm

lol

wait a second;

(all the semicolons)

no that too didn't work

I tried adding and subtracting the same thing which made a mess of it

:|

I got one easy integral and one that's the reciprocal of this original

but that doesn't quite look as if that were the answer :(

Ask WolframAlpha

ok

negative tanh inverse of the denominator ? How :|

:|

should I try integrating by parts?

No! Are you crazy!

xD

but I feel somehow it's just a substituion

It should just be a substitution problem

right

Perhaps you differentiated wrong

I think WA agrees with your result

oh maybe

are you sure?

idk

Dumb answers of mine :|