- #1
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Let's say i feed Maple with this baby
[tex] \int_{0}^{\infty }\frac{x\ln x}{1+e^{x}}dx [/tex]
After some thinking he gives me
[tex] \int_{0}^{\infty }\frac{x\ln x}{1+e^{x}}dx = \frac{1}{12}\left( \ln 2\right) \pi ^{2}+\frac{1}{12}\pi ^{2}-\frac{1}{12}\pi ^{2}\gamma+\frac{1}{2}\zeta \left( 1,2\right) [/tex]
I say OK, even though i don't understand who [itex] \zeta \left( 1,2\right) [/itex] is. Is it Hurwitz zeta function ? If so, then it should diverge... Anyway, the same software tells me that
[tex] \zeta \left( 1,2\right) = -.\,93754\,82543 [/tex]
So i say OK again, even though I'm not satisfied with what zeta stands for...
But that's not the issue. The issue is that I'm putting my head to work to do that integral and i get
[tex] \int_{0}^{\infty }\frac{x\ln x}{1+e^{x}}dx = \frac{\pi ^{2}}{12}\left( 1-\gamma \right) -\sum_{k=0}^{\infty }\left[ \left( -\right) ^{k}\frac{\ln \left( k+1\right) }{\left( k+1\right) ^{2}}\right] [/tex]
So I'm thinking that the nasty sum should equal the 2 terms above, the one with the natural logarithm and the other with the dubious zeta function. But when i feed Maple with the sum, he doesn't return anything...
So what the heck? Dumb software, i figure... He woudn't give me not even an aproximate numerical value of the sum...
So the question is: any ideas on how to evaluate the sum...?
Daniel.
[tex] \int_{0}^{\infty }\frac{x\ln x}{1+e^{x}}dx [/tex]
After some thinking he gives me
[tex] \int_{0}^{\infty }\frac{x\ln x}{1+e^{x}}dx = \frac{1}{12}\left( \ln 2\right) \pi ^{2}+\frac{1}{12}\pi ^{2}-\frac{1}{12}\pi ^{2}\gamma+\frac{1}{2}\zeta \left( 1,2\right) [/tex]
I say OK, even though i don't understand who [itex] \zeta \left( 1,2\right) [/itex] is. Is it Hurwitz zeta function ? If so, then it should diverge... Anyway, the same software tells me that
[tex] \zeta \left( 1,2\right) = -.\,93754\,82543 [/tex]
So i say OK again, even though I'm not satisfied with what zeta stands for...
But that's not the issue. The issue is that I'm putting my head to work to do that integral and i get
[tex] \int_{0}^{\infty }\frac{x\ln x}{1+e^{x}}dx = \frac{\pi ^{2}}{12}\left( 1-\gamma \right) -\sum_{k=0}^{\infty }\left[ \left( -\right) ^{k}\frac{\ln \left( k+1\right) }{\left( k+1\right) ^{2}}\right] [/tex]
So I'm thinking that the nasty sum should equal the 2 terms above, the one with the natural logarithm and the other with the dubious zeta function. But when i feed Maple with the sum, he doesn't return anything...
So what the heck? Dumb software, i figure... He woudn't give me not even an aproximate numerical value of the sum...
So the question is: any ideas on how to evaluate the sum...?
Daniel.