Question

How to show that $\int_0^{+\infty} e^{-u} \frac{\sin(xu)}{u}\operatorname{d}u=\arctan(x)$?

score 87 · Answer 1 · 2019-03-05 23:40:08

87

Views

How to show that $\int_0^{+\infty} e^{-u} \frac{\sin(xu)}{u}\operatorname{d}u=\arctan(x)$?

Published on 05 Mar 2019 - 23:40

score 64 · Answer 2 · 2026-03-30 05:00:28

64

Views

What does $\asymp$ with regards to improper integrals?

Published on 30 Mar 2026 - 5:00

score 209 · Answer 3 · 2019-03-06 05:42:48

209

Views

Is $\int_{0^+}^{+\infty}\frac{e^{-x}}{\sqrt{x}}dx$ convergent or divergent?

Published on 06 Mar 2019 - 5:42

score 317 · Answer 4 · 2019-03-06 07:05:14

317

Views

Integral ${\frac{1}{\pi^2}} \int_{0}^{\infty}\frac{{(\ln{x}})^2}{\sqrt{x}{(1-x)^2}} \mathrm d x$

Published on 06 Mar 2019 - 7:05

score 90 · Answer 5 · 2019-03-06 10:40:32

90

Views

Integral of odd function gives complex value

Published on 06 Mar 2019 - 10:40

score 103 · Answer 6 · 2019-03-06 16:07:21

103

Views

How to deal with improper integrals that result in indeterminate forms?

Published on 06 Mar 2019 - 16:07

score 100 · Answer 7 · 2026-03-27 07:57:21

100

Views

Integrals involving multi-valued functions

Published on 27 Mar 2026 - 7:57

score 127 · Answer 8 · 2019-03-07 03:23:41

127

Views

Area beneath $y=x$ from $-\infty$ to $\infty$

Published on 07 Mar 2019 - 3:23

score 69 · Answer 9 · 2026-03-26 12:37:16

69

Views

Solving $\int_0^x\int_u^{u+1}\frac1{\ln t}\,dt\,du=0$ (an extension to the Ramanujan-Soldner constant)

Published on 26 Mar 2026 - 12:37

score 82 · Answer 10 · 2019-03-08 10:12:06

82

Views

Improper integrals, one-sided limits and trigonometric functions

Published on 08 Mar 2019 - 10:12

score 67 · Answer 11 · 2026-03-27 22:51:10

67

Views

Can some improper integrals be directly evaluated?

Published on 27 Mar 2026 - 22:51

score 54 · Answer 12 · 2019-03-09 16:00:01

54

Views

Antiderivative of $\frac{(x+2)^2}{\sqrt{1-\frac{1}{x^2}}}$

Published on 09 Mar 2019 - 16:00

score 64 · Answer 13 · 2019-03-10 13:07:36

64

Views

Does $\int_{0}^{\pi ^2} \frac{dx}{1 - \cos \sqrt{x}}$ converge or diverge?

Published on 10 Mar 2019 - 13:07

score 72 · Answer 14 · 2019-03-10 16:31:16

72

Views

Improper Integration Technique

Published on 10 Mar 2019 - 16:31

score 62 · Answer 15 · 2026-03-29 16:29:07

62

Views

Improper integral convergence for different powers of a fraction

Published on 29 Mar 2026 - 16:29

List Question

How to show that $\int_0^{+\infty} e^{-u} \frac{\sin(xu)}{u}\operatorname{d}u=\arctan(x)$?

What does $\asymp$ with regards to improper integrals?

Is $\int_{0^+}^{+\infty}\frac{e^{-x}}{\sqrt{x}}dx$ convergent or divergent?

Integral ${\frac{1}{\pi^2}} \int_{0}^{\infty}\frac{{(\ln{x}})^2}{\sqrt{x}{(1-x)^2}} \mathrm d x$

Integral of odd function gives complex value

How to deal with improper integrals that result in indeterminate forms?

Integrals involving multi-valued functions

Area beneath $y=x$ from $-\infty$ to $\infty$

Solving $\int_0^x\int_u^{u+1}\frac1{\ln t}\,dt\,du=0$ (an extension to the Ramanujan-Soldner constant)

Improper integrals, one-sided limits and trigonometric functions

Can some improper integrals be directly evaluated?

Antiderivative of $\frac{(x+2)^2}{\sqrt{1-\frac{1}{x^2}}}$

Does $\int_{0}^{\pi ^2} \frac{dx}{1 - \cos \sqrt{x}}$ converge or diverge?

Improper Integration Technique

Improper integral convergence for different powers of a fraction

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