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9h
comment integrate this double integral by any method you can.
Possible duplicate: math.stackexchange.com/questions/1019234/…
13h
asked Definite integral of arcsine over square-root of quadratic
23h
comment Definite integral involving Legendre Polynomial
May we assume $\ell$ is a non-negative integer?
May
22
comment Definite integral of polynomial times exponential times hypergeometric function of imaginary argument
In the interest of housekeeping, could please clarify what parts of the problem you still need answered?
May
22
answered A series involving harmonic numbers
May
16
comment What is the triple factorial of a negative number, e.g., $-2$?
This is probably the first question here with a title ending in triple exclamation points that I didn't automatically downvote. :)
May
15
comment Computing $\int_{0}^{\pi/2}\cos(x)\ln(\tan(x))dx$
I'm not the downvoter, but my guess would be that your question was downvoted on grounds of being a duplicate (though if that's the case, link to duplicate question really should have been provided).
May
9
comment Solve $\int_{0}^{1} \log(x)\log(1-x) dx$ without convolution
@AnjanDebnath Yes. One way to demonstrate this is by Taylor expanding the $\ln{(x)}$ factor about $x=1$. You could also use l'Hoptial's rule of course.
May
9
answered Solve $\int_{0}^{1} \log(x)\log(1-x) dx$ without convolution
May
9
answered Calculating $\int_0^{\infty } \frac{\log (v+1)}{\sqrt{(v+1)^2+1} \sqrt{(v+1)^2+4 \sqrt{(v+1)^2+1} (v+1)+4}} \, dv$
May
7
awarded  Revival
May
6
comment Calculating $\int_0^{\infty } \frac{\log (v+1)}{\sqrt{(v+1)^2+1} \sqrt{(v+1)^2+4 \sqrt{(v+1)^2+1} (v+1)+4}} \, dv$
Heh. I initially thought I was making quick work of this integral, but I'm fairly sure what I obtained is going to leave me with a ${_4F_3}$ term that I won't be able to simplify. Since I take it your ultimate aim is to somehow use this to eliminate a ${_3F_2}$ in your other question, I have a feeling this might be heading in the wrong direction...
May
6
answered Integration of the incomplete beta function
May
6
comment Dealing with an integral: can we go any farther?
@martycohen I have. The first thing I checked was to make sure my closed form returned the same known value at $a=0$ that Lucian mentioned in his comment above. and that it had the correct limiting behavior $I(a)\to0$ as $a\to1^{-}$. Both of these checked out. I also checked for numerical consistency at a few rational values of $a$ between $0$ and $1$ to about six decimal places using WolframAlpha, and there was no disparity.
May
6
revised Dealing with an integral: can we go any farther?
minor changes
May
5
answered Dealing with an integral: can we go any farther?
May
4
awarded  Revival
Apr
26
answered Find the closed form of the digamma related series
Apr
26
awarded  Revival
Apr
25
answered What is a $0\times0$ or $0\times3$ matrix?