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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 Post subject: Re: $\sum_{n=2}^{\infty}(-1)^n \frac{\ln n}{n} =?$ 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: $$\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^{a}}=\frac{1}{\Gamma(a)}\sum_{n=1}^{\infty}\cos(nx)\int_{0}^{\infty}e^{-nt}t^{a-... Forum: Calculus Topic: how about this log generalization?  Post subject: how about this log generalization? Posted: Sat Mar 12, 2016 2:17 am Replies: 1 Views: 541  Derive a general closed form for:$$\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 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  Sort by:
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