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1d
revised How do you prove this integral involving the Glaisher–Kinkelin constant
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1d
revised How do you prove this integral involving the Glaisher–Kinkelin constant
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1d
comment Evaluating $\int_0^\pi\arctan\left(\frac{\ln\sin x}{x}\right)\mathrm{d}x$
I actually upvoted your answer yesterday. But I thought I'd leave a comment today since it's not getting the upvotes it deserves.
1d
comment Evaluating $\int_0^\pi\arctan\left(\frac{\ln\sin x}{x}\right)\mathrm{d}x$
I'm surprised your answer only has 3 upvotes (including mine). It deserves a lot more.
1d
revised The log gamma integral $\int_{0}^{z} \log \Gamma (x) \ \mathrm dx$
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1d
revised How do you prove this integral involving the Glaisher–Kinkelin constant
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1d
answered How do you prove this integral involving the Glaisher–Kinkelin constant
2d
revised Generalised Integral $I_n=\displaystyle \int_0^{\pi/2} \frac{x^n}{\sin ^n x} \ \mathrm{d}x, \quad n\in \mathbb{Z}^+.$
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2d
revised Equivalence of integral and sum
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2d
revised Generalised Integral $I_n=\displaystyle \int_0^{\pi/2} \frac{x^n}{\sin ^n x} \ \mathrm{d}x, \quad n\in \mathbb{Z}^+.$
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2d
answered Generalised Integral $I_n=\displaystyle \int_0^{\pi/2} \frac{x^n}{\sin ^n x} \ \mathrm{d}x, \quad n\in \mathbb{Z}^+.$
Jul
28
revised Do these integrals have a closed form? $I_1 = \int_{-\infty }^{\infty } \frac{\sin (x)}{x \cosh (x)} \, dx$
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Jul
28
revised Do these integrals have a closed form? $I_1 = \int_{-\infty }^{\infty } \frac{\sin (x)}{x \cosh (x)} \, dx$
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Jul
26
comment Evaluate $\int\left({\frac{\arctan x}{\arctan x-x}}\right)^2 \,dx$
math.stackexchange.com/questions/79074/…
Jul
25
comment Prove ${\large\int}_0^\infty\frac{\ln x}{\sqrt{x}\ \sqrt{x+1}\ \sqrt{2x+1}}dx\stackrel?=\frac{\pi^{3/2}\,\ln2}{2^{3/2}\Gamma^2\left(\tfrac34\right)}$
Very nice answer. +1 After about 30 minutes of getting nowhere, I gave up.
Jul
25
revised Prove this polylogarithmic integral has the stated closed form value
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Jul
25
revised Prove this polylogarithmic integral has the stated closed form value
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Jul
25
answered Prove this polylogarithmic integral has the stated closed form value
Jul
24
revised Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$
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Jul
24
comment Evaluating $\int_{0}^{\infty} \frac{x^{3}- \sin^{3}(x)}{x^{5}} \ dx $ using contour integration
@robjohn Isn't the difference that you shifted the contour while I moved the pole?