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Forum: Calculus Topic: Logarithmic integral 
galactus 
Posted: Thu Jul 14, 2016 6:26 am


Replies: 2 Views: 552

The integral can be rewritten 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 
galactus 
Posted: Sat Jul 09, 2016 5:41 am


Replies: 0 Views: 320

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



Forum: Calculus Topic: $\sum_{n=2}^{\infty}(1)^n \frac{\ln n}{n} =?$ 
galactus 
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: $$\frac{1}{n^{a}}=\frac{1}{\Gamma(a)}\int_{0}^{\infty}e^{nt}t^{a1}dt$$ 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? 
galactus 
Posted: Sat Mar 12, 2016 2:17 am


Replies: 1 Views: 541

Derive a general closed form for: $$\int_{0}^{\infty}\left(\frac{\log(x)}{x1}\right)^{n}dx$$ $$=n\sum_{m=1}^{\infty}\frac{1}{m^{n}}\left(\prod_{k=1}^{n2}(m+k)+\prod_{k=1}^{n2}(mk)\right)$$ 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 
galactus 
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 
galactus 
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 
galactus 
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 
galactus 
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 
galactus 
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 
galactus 
Posted: Sun Feb 14, 2016 11:20 pm


Replies: 3 Views: 1044

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


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