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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 ...

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

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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