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Seeking for $\alpha\sum_{k=1}^{\infty}{(-1)^{k-1}\over k(e^{k\pi}-1)}+\beta\sum_{k=1}^{\infty}{(-1)^{k-1}\over k(e^{k\pi}+1)}=f(\alpha,\beta)$

score 122 · Answer 1 · 2017-06-20 18:58:45

122

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Seeking for $\alpha\sum_{k=1}^{\infty}{(-1)^{k-1}\over k(e^{k\pi}-1)}+\beta\sum_{k=1}^{\infty}{(-1)^{k-1}\over k(e^{k\pi}+1)}=f(\alpha,\beta)$

Published on 20 Jun 2017 - 18:58

score 1.2k · Answer 2 · 2026-03-28 02:06:39

1.2k

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374

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score 776 · Answer 4 · 2026-03-28 03:27:39

776

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score 406 · Answer 5 · 2026-04-06 05:53:39

406

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127

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83

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score 933 · Answer 8 · 2026-03-30 02:52:21

933

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408

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score 440 · Answer 10 · 2026-04-06 05:58:46

440

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Published on 06 Apr 2026 - 5:58

score 686 · Answer 11 · 2017-07-08 17:25:53

686

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score 56 · Answer 12 · 2017-07-09 09:14:51

56

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1k

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score 413 · Answer 14 · 2026-04-07 16:51:39

413

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score 309 · Answer 15 · 2026-04-08 02:33:35

309

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Published on 08 Apr 2026 - 2:33

List Question

Seeking for $\alpha\sum_{k=1}^{\infty}{(-1)^{k-1}\over k(e^{k\pi}-1)}+\beta\sum_{k=1}^{\infty}{(-1)^{k-1}\over k(e^{k\pi}+1)}=f(\alpha,\beta)$

Prove that $\int_0^\infty \frac{1+2\cos x+x\sin x}{1+2x\sin x +x^2}dx=\frac{\pi}{1+\Omega}$ where $\Omega e^\Omega=1$

Pick's theorem for a triangle

Closed form for ${\large\int}_0^1\frac{\ln^4(1+x)\ln x}x \, dx$

Closed form of a sequence defined by two indices and a recursion formula

General term and closed form of a fraction of fraction of a fraction of a ...

Finding closed form of a summation $\sum_{i = \lfloor{\frac{n}{2}}\rfloor}^{n}i$

Maximum Likelihood Estimator (MLE ) for the Gaussian-noise simple linear regression model

Series involving Bernoulli Numbers

Is there a closed form formula for the recursive sequence: $x_n = x_{n-1} + \alpha\sqrt{x_{n-1}}$

Show this $\int_0^\infty \frac{t\ln(2\sinh t)}{\left(3t^2+\ln^2(2\sinh t)\right)^2}~dt=0$

Closed-form of an two-dimensional integral

High-order antiderivatives of the Weierstrass P-function

How to evaluate a Lambert-type series in finite terms?

How to find solutions to equation $f'(x)+f(x)-Kf(-x)=0$

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