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Jan
23
comment Relations between hypergeometric functions
@Startwearingpurple: I was hoping the 'some other hypergeometric' would actually be a trivial hypergeometric in the sense that its arguments are such that the hypergeometric turns out to be just $z$ to some power or something at least simpler than a hypergeometric.
Jan
16
asked Relations between hypergeometric functions
Jan
14
accepted Multiplying two summations together exactly.
Dec
27
accepted Computing a double integral
Dec
27
accepted Generators of a semi simple lie algebra must be traceless
Dec
27
accepted Chain rule and partial derivatives for multivariable function
Dec
8
accepted Simple inequalities
Dec
8
comment Simple inequalities
Ok no problem! thanks anyway
Dec
8
comment Simple inequalities
Ah sorry, it was under 'Generic general cases', the very last one in that list.
Dec
8
comment Simple inequalities
Ok, many thanks, it is what I thought. Kind of unrelated to the question, but in this website of hypergeometric identies, functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/…, it relates the hypergeometric 2_F_1(a,b,c,z) = U(a,b,a+b-c+1, 1-z). Would you know what U stands for here? The reason I wanted |1-z|<1 (and that the prof incorrectly concluded) was so that I could make a transformation to a converging hypergeometric.
Dec
8
revised Simple inequalities
edited title
Dec
8
asked Simple inequalities
Dec
7
comment Evaluation of double integral using substitution
? Is an integral like $$\int_0^1 dw \frac{w^{\alpha+1} \ln(b+1-w)}{1+b(1+w)}$$ really non trivial? I mean I think it should be hypergeometric but I can't see how to transform it into one
Dec
7
revised Integral of the principal value of a hypergeometric function
added 282 characters in body
Dec
7
asked Integral of the principal value of a hypergeometric function
Dec
1
comment Evaluation of double integral using substitution
Do you think integration by parts would help with the above?
Dec
1
comment Non trivial sum involving Gamma functions and hypergeometric
@Lucian: :P, basically the integral I am trying to evaluate is the following $$\int_0^1 dz \int_0^1 du \frac{u^{1+\alpha} (1-z)^{\alpha}z}{(b+z)(1-uz)},$$ for $b$ and $\alpha$ real constants not necessarily integers. In the above I reexpressed the coupled term (1-uz) as a geometric series. I've tried multiple things, like substitutions and partial fractions but all seem doomed to fail. Would you have any further ideas?
Nov
30
comment Evaluation of double integral using substitution
A partial fraction decomposition does indeed give logs and polynomials but the resulting integral is $$\int_0^1 dw \frac{w^{\alpha+1}}{1+b(1+w)} \left(1-w-b\ln(1+1/b-w/b)-2\ln w\right),$$ The first two terms look like hypergeometrics but the last two terms I am not sure.
Nov
30
comment Evaluation of double integral using substitution
@zahbaz: I got a factor of $w/(1-z)$ when I subbed in for $u$ on the numerator but this was cancelled by a $(1-z)$ coming from replacing $1-uz$ in terms of $w$ and $z$. (as far as I can see). And yes, they are just real constants.
Nov
30
comment Evaluation of double integral using substitution
Hi @tired, hmm I don't see why the inner integral is elementary now, doesn't the coupled term $wz$ complicate things?