Search found 56 matches
- Thu Jul 14, 2016 6:26 am
- Forum: Calculus
- Topic: Logarithmic integral
- Replies: 2
- Views: 2851
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...
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
- Thu Mar 24, 2016 9:58 pm
- Forum: Calculus
- Topic: $\sum_{n=2}^{\infty}(-1)^n \frac{\ln n}{n} =?$
- Replies: 2
- Views: 3580
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: $$\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^{a}}=\frac{1}{\Gamma(a)}\sum_{n=1}^{\infty}\cos(nx)\int_{0}^{\infty}e^{-nt}t^{a-...
- Sat Mar 12, 2016 2:17 am
- Forum: Calculus
- Topic: how about this log generalization?
- Replies: 1
- Views: 1959
how about this log generalization?
Derive a general closed form for: $$\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}^{\...
- Sun Mar 06, 2016 5:48 pm
- Forum: Calculus
- Topic: A sine series
- Replies: 2
- Views: 2409
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 $$\frac{-\pi^{3}\sqrt{2}}{36}$$ $$f(z)=\frac{\pi cot(\pi z)csc(\pi \sqrt{2}z)}{z^{3}}$$...
- Tue Mar 01, 2016 10:27 pm
- Forum: Calculus
- Topic: Nice integral problem
- Replies: 7
- Views: 5327
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...
- Sun Feb 28, 2016 2:15 pm
- Forum: Calculus
- Topic: A definite integral
- Replies: 2
- Views: 2367
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 $u=\ln(t), \;\ dv=\frac{1}{t^{2}+a}dt$ $$\underbrace{\frac{\ln(a)\tan^{-1}(\sqrt{a})}{\sqrt{a}}}_{\text{1}}-\frac{1}{\sqrt{...
- Sat Feb 27, 2016 4:20 pm
- Forum: Calculus
- Topic: A tough series
- Replies: 1
- Views: 1949
Re: A tough series
I deleted what I posted because I was dissatisfied with it. There were convergence issues. Using digamma and the arctan series, I found the sum is equal to the integral: $$\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...
- Fri Feb 26, 2016 1:15 pm
- Forum: Calculus
- Topic: A Laplace transform
- Replies: 2
- Views: 2543
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(...
- Sun Feb 14, 2016 11:20 pm
- Forum: Calculus
- Topic: A tough product
- Replies: 3
- Views: 4029
Re: A tough product
Wow, rd, you're just showin' off now