Question

Little curiosity $\Re\Big(\int_{0}^{1}\frac{x\ln(x)}{1+e^{ix}}dx\Big)=-\frac{1}{8}$

score 83 · Answer 1 · 2026-03-30 05:26:56

83

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Little curiosity $\Re\Big(\int_{0}^{1}\frac{x\ln(x)}{1+e^{ix}}dx\Big)=-\frac{1}{8}$

Published on 30 Mar 2026 - 5:26

score 56 · Answer 2 · 2026-03-25 04:40:26

56

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Showing that the difference of these 2 meromorphic functions are bounded

Published on 25 Mar 2026 - 4:40

score 77 · Answer 3 · 2020-04-18 18:06:50

77

Views

Integral using residues

Published on 18 Apr 2020 - 18:06

score 612 · Answer 4 · 2026-02-23 03:27:50

612

Views

Evaluate $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2}$ with Poisson summation formula

Published on 23 Feb 2026 - 3:27

score 81 · Answer 5 · 2026-03-28 14:44:15

81

Views

Residue of $z^4e^\frac{1}{z-1}$ at $z=1$

Published on 28 Mar 2026 - 14:44

score 28 · Answer 6 · 2026-03-28 14:44:05

28

Views

Why does it seem like we are selectively cancelling $(z-z_0)$ terms in the Laurent series in the limit $z\rightarrow z_0$?

Published on 28 Mar 2026 - 14:44

score 122 · Answer 7 · 2020-04-22 15:11:55

122

Views

Integrating $\arccos(x)$ with the residue theorem

Published on 22 Apr 2020 - 15:11

score 156 · Answer 8 · 2020-04-23 03:05:20

156

Views

Compute $ \oint_{C_{o}} \frac{1}{\sin \left(\frac{1}{z}\right)} d z-\oint_{C_{i}} \frac{1}{\sin \left(\frac{1}{z}\right)} \mathrm d z $

Published on 23 Apr 2020 - 3:05

score 236 · Answer 9 · 2026-03-28 14:37:12

236

Views

Contour Integral from Laurent Series

Published on 28 Mar 2026 - 14:37

score 28 · Answer 10 · 2020-04-24 03:18:00

28

Views

problem with final part of this residue integral.

Published on 24 Apr 2020 - 3:18

score 47 · Answer 11 · 2020-04-24 23:33:51

47

Views

what is my mistake in estimating CR integral by Jordan s lemma

Published on 24 Apr 2020 - 23:33

score 34 · Answer 12 · 2020-04-25 18:06:01

34

Views

$\operatorname{Res}(f,-i)$ for $f(z) = \frac{1}{z} + \exp\left(\frac{z-i}{z^2+1} \right)$ at $z = -i$

Published on 25 Apr 2020 - 18:06

score 123 · Answer 13 · 2020-04-26 22:46:54

123

Views

$\int_0^\infty \frac{1-\cos(x)}{x^2}$.

Published on 26 Apr 2020 - 22:46

score 43 · Answer 14 · 2026-03-28 14:44:33

43

Views

theory of Laurent expansions,

Published on 28 Mar 2026 - 14:44

score 100 · Answer 15 · 2020-04-28 12:53:03

100

Views

Finding residue of complex function, the result is different when using Laurent series and residue theorem.

Published on 28 Apr 2020 - 12:53

List Question

Little curiosity $\Re\Big(\int_{0}^{1}\frac{x\ln(x)}{1+e^{ix}}dx\Big)=-\frac{1}{8}$

Showing that the difference of these 2 meromorphic functions are bounded

Integral using residues

Evaluate $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2}$ with Poisson summation formula

Residue of $z^4e^\frac{1}{z-1}$ at $z=1$

Why does it seem like we are selectively cancelling $(z-z_0)$ terms in the Laurent series in the limit $z\rightarrow z_0$?

Integrating $\arccos(x)$ with the residue theorem

Compute $ \oint_{C_{o}} \frac{1}{\sin \left(\frac{1}{z}\right)} d z-\oint_{C_{i}} \frac{1}{\sin \left(\frac{1}{z}\right)} \mathrm d z $

Contour Integral from Laurent Series

problem with final part of this residue integral.

what is my mistake in estimating CR integral by Jordan s lemma

$\operatorname{Res}(f,-i)$ for $f(z) = \frac{1}{z} + \exp\left(\frac{z-i}{z^2+1} \right)$ at $z = -i$

$\int_0^\infty \frac{1-\cos(x)}{x^2}$.

theory of Laurent expansions,

Finding residue of complex function, the result is different when using Laurent series and residue theorem.

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