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 Forum: Calculus   Topic: Logarithmic integral

 Post subject: Re: Logarithmic integral
Posted: Thu Jul 14, 2016 6:26 am 

Replies: 2
Views: 552


The integral can be re-written as: $$\int_{0}^{1}\frac{\ln^2 x}{x^2-\sqrt{2}x+1}\,dx=\frac{1}{2}\int_{0}^{\infty}\frac{\ln^2 x}{x^2-\sqrt{2}x+1}\,dx$$ Considering \( \displaystyle \oint _{\gamma}\frac{\ln^3 z}{z^2-\sqrt{2}z+1}\,dz \) whereas \( \gamma \) is a circle contour with a branch along the p...

 Forum: Calculus   Topic: Series

 Post subject: Series
Posted: Sat Jul 09, 2016 5:41 am 

Replies: 0
Views: 320


Prove that: \( \displaystyle \sum_{n=1}^{\infty }\binom{2n}{n}^2n\frac{1}{2^{5n}}H_n=1-\frac{\Gamma ^2\left ( \frac{3}{4} \right )}{\pi\sqrt{\pi}}\left ( \frac{\pi}{2}+2\ln 2 \right ) \).


NOTE
I don't have a solution but I'd would like to see one, unless it involves hypergeometric series.

 Forum: Calculus   Topic: $\sum_{n=2}^{\infty}(-1)^n \frac{\ln n}{n} =?$

Posted: Thu Mar 24, 2016 9:58 pm 

Replies: 2
Views: 812


I am gonna begin this way. Of course, I am sure there are many ways to begin. Start with the classic: $$\frac{1}{n^{a}}=\frac{1}{\Gamma(a)}\int_{0}^{\infty}e^{-nt}t^{a-1}dt$$ Then: $$\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^{a}}=\frac{1}{\Gamma(a)}\sum_{n=1}^{\infty}\cos(nx)\int_{0}^{\infty}e^{-nt}t^{a-...

 Forum: Calculus   Topic: how about this log generalization?

Posted: Sat Mar 12, 2016 2:17 am 

Replies: 1
Views: 541


Derive a general closed form for: $$\int_{0}^{\infty}\left(\frac{\log(x)}{x-1}\right)^{n}dx$$ $$=n\sum_{m=1}^{\infty}\frac{1}{m^{n}}\left(\prod_{k=1}^{n-2}(m+k)+\prod_{k=1}^{n-2}(m-k)\right)$$ or whatever form you come up with that gives the correct solution. e.g. Let $n=6$ and we get: $$\int_{0}^{\...

 Forum: Calculus   Topic: A sine series

 Post subject: Re: A sine series
Posted: Sun Mar 06, 2016 5:48 pm 

Replies: 2
Views: 648


Yes, I see now. This is one of those tricky SOB's. :roll: :oops: I used the wrong f(z) a while ago anyway. I ran it through Maple for a check and it returns the correct result that you posted. The residue at 0 is $$\frac{-\pi^{3}\sqrt{2}}{36}$$ $$f(z)=\frac{\pi cot(\pi z)csc(\pi \sqrt{2}z)}{z^{3}}$$...

 Forum: Calculus   Topic: Nice integral problem

 Post subject: Re: Nice integral problem
Posted: Tue Mar 01, 2016 10:27 pm 

Replies: 7
Views: 1457


Yeah, T, I would say that is probably the case. There are some fun integrals associated with this one. One may derive the integral you mention by using Zeta, so it is more than likely safe to assume we can continue it to $0<s<1$. Like you said, showing the one you mention can be done using geometric...

 Forum: Calculus   Topic: A definite integral

 Post subject: Re: A definite integral
Posted: Sun Feb 28, 2016 2:15 pm 

Replies: 2
Views: 621


I think I may have this one. $$\int_{1}^{a^{2}}\frac{\ln(x)}{\sqrt{x}(x+a)}dx$$ Let $x=t^{2}, \;\ 2dt=\frac{1}{\sqrt{x}}dx$ $$4\int_{1}^{a}\frac{\ln(t)}{t^{2}+a}dt$$ Parts, Let $u=\ln(t), \;\ dv=\frac{1}{t^{2}+a}dt$ $$\underbrace{\frac{\ln(a)\tan^{-1}(\sqrt{a})}{\sqrt{a}}}_{\text{1}}-\frac{1}{\sqrt{...

 Forum: Calculus   Topic: A tough series

 Post subject: Re: A tough series
Posted: Sat Feb 27, 2016 4:20 pm 

Replies: 1
Views: 569


I deleted what I posted because I was dissatisfied with it. There were convergence issues. Using digamma and the arctan series, I found the sum is equal to the integral: $$\int_{0}^{1}\frac{x(\tan^{-1}(x)-\tan^{-1}(x^{5}))}{x^{4}-1}dx=\int_{0}^{1}\frac{x\tan^{-1}\left(\frac{x(x^{2}-1)}{x^{4}-x^{2}+1...

 Forum: Calculus   Topic: A Laplace transform

 Post subject: Re: A Laplace transform
Posted: Fri Feb 26, 2016 1:15 pm 

Replies: 2
Views: 702


Clever and efficient, RD. :clap2: Well, I scratched around with this one a little. I did not want to do the same exact thing as R. I do not know how Seraphim done it, so here goes: $$\int_{0}^{\infty}e^{-ax}\sin(x)\sin(\sqrt{x})dx$$ Use identities for both sin terms, one the Taylor series for $\sin(...

 Forum: Calculus   Topic: A tough product

 Post subject: Re: A tough product
Posted: Sun Feb 14, 2016 11:20 pm 

Replies: 3
Views: 1044


Wow, rd, you're just showin' off now :clap2:
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