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10h
revised Calculating in closed form another digamma alternating series
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10h
comment Calculating in closed form another digamma alternating series
Maybe. But I like integrals.
10h
revised Calculating in closed form another digamma alternating series
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10h
answered Calculating in closed form another digamma alternating series
1d
comment Calculating $\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$
The inner integral of the double integral is easily done via the residue theorem.
Aug
23
comment Estimate integral $\,\displaystyle\int_{0}^{\infty}\operatorname{sech}\left(\varepsilon x\right)\cos\left(kx\right)\,dx,\,$ with $\,k,\varepsilon>0$
Yes ${}{}{}{}{}{}{}$
Aug
23
comment Estimate integral $\,\displaystyle\int_{0}^{\infty}\operatorname{sech}\left(\varepsilon x\right)\cos\left(kx\right)\,dx,\,$ with $\,k,\varepsilon>0$
I don't think you need an asymptotic estimate of this integral. It is essentially a Fourier transform of the Sech function, which is another Sech function.
Aug
21
comment Approximate an integral with Bessel functions
Yes, but even so, it may be profitable to consider limits of $n$ both small and large. In that way, something like a two-point Pade approximant may be constructed.
Aug
21
answered Show that $\displaystyle{\int_{0}^{\infty}\!\frac{x^{a}}{x(x+1)}~\mathrm{d}x=\frac{\pi}{\sin(\pi a)}}$
Aug
21
comment Evaluate $\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)\;dx$
And it is appreciated. I just wanted to point that out. No reflection on your good work.
Aug
21
revised Evaluate $\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)\;dx$
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Aug
20
comment Evaluate $\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)\;dx$
@cody was that wise young man.
Aug
20
comment Evaluate $\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)\;dx$
As a wise young man once said, there's more than one way to skin an integral.
Aug
20
revised Evaluate $\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)\;dx$
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Aug
20
answered Evaluate $\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)\;dx$
Aug
20
revised $\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$
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Aug
20
revised $\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$
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Aug
20
comment $\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$
@StevenStadnicki: Thanks, and yes.
Aug
20
answered $\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$
Aug
20
comment calculating you weight if you $100\text{Km}$ above the earth etc.
There's a small effect that is, at the equator, proportional to the frequency of rotation squared, which is $m R_e \omega^2$. The earth's rotation being 10h means the rotation frequency increases, so the weight increases.