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by galactus
Thu Jul 14, 2016 6:26 am
Forum: Calculus
Topic: Logarithmic integral
Replies: 2
Views: 728

Re: Logarithmic integral

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...
by galactus
Sat Jul 09, 2016 5:41 am
Forum: Calculus
Topic: Series
Replies: 0
Views: 429

Series

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.
by galactus
Thu Mar 24, 2016 9:58 pm
Forum: Calculus
Topic: $\sum_{n=2}^{\infty}(-1)^n \frac{\ln n}{n} =?$
Replies: 2
Views: 987

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

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-...
by galactus
Sat Mar 12, 2016 2:17 am
Forum: Calculus
Topic: how about this log generalization?
Replies: 1
Views: 664

how about this log generalization?

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}^{\...
by galactus
Sun Mar 06, 2016 5:48 pm
Forum: Calculus
Topic: A sine series
Replies: 2
Views: 765

Re: A sine series

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 $$\frac{-\pi^{3}\sqrt{2}}{36}$$ $$f(z)=\frac{\pi cot(\pi z)csc(\pi \sqrt{2}z)}{z^{3}}$$...
by galactus
Tue Mar 01, 2016 10:27 pm
Forum: Calculus
Topic: Nice integral problem
Replies: 7
Views: 1724

Re: Nice integral problem

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...
by galactus
Sun Feb 28, 2016 2:15 pm
Forum: Calculus
Topic: A definite integral
Replies: 2
Views: 733

Re: A definite integral

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{...
by galactus
Sat Feb 27, 2016 4:20 pm
Forum: Calculus
Topic: A tough series
Replies: 1
Views: 669

Re: A tough series

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...
by galactus
Fri Feb 26, 2016 1:15 pm
Forum: Calculus
Topic: A Laplace transform
Replies: 2
Views: 802

Re: A Laplace transform

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(...
by galactus
Sun Feb 14, 2016 11:20 pm
Forum: Calculus
Topic: A tough product
Replies: 3
Views: 1275

Re: A tough product

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