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Show that:

$$\int_{0}^{\pi/2}\left[\left(\frac{3}{x^{2}}-1\right)\sin(x)-\frac{3}{x}\cos(x)\right]^{2}dx=\frac{\pi}{4}-\frac{24}{\pi^{3}}$$

Wow!, looks like a tricky one with the integration in the exponent. But, write as: $$\int_{0}^{\pi/4}\int_{\pi/2}^{\pi}\frac{\sin^{y-2}\left(\frac{\pi}{4}-x\right)}{\sin^{y+2}\left(\frac{\pi}{4}+x\right)}dydx$$ Switch integral signs: $$\int_{\pi/2}^{\pi}\int_{0}^{\pi/4}\frac{\sin^{y-2}\left(\frac{\p...

Wow, very nice rd

Cool, T

$$\sum_{n=0}^{\infty}\binom{3n}{n}\left(2/27\right)^{n}$$ Just like $\sum_{n=0}^{\infty}\binom{2n}{n}x^{n}=\frac{1}{\sqrt{1-4x}}$ there is a closed form gen. func. for $\binom{3n}{n}$ series as well. Though, I have to admit, I have not derived it as yet. Maybe we'll have to do that. begin with: $$=\...

Nice use of the partial sum/stirling thing there at the end RD. I like.

My solution is not very glamorous. Write the integrand as: $$1/2\int_{0}^{\infty}\left(\frac{1}{\sinh(x)}-\frac{1}{x}\right)\cdot \frac{1}{x}dx$$ Consider the Dirichlet Lambda Function. For $s>1$, the DLF is defined as : $$\lambda(s) = \sum_{k=0}^{\infty}\frac{1}{(2k+1)^{s}}=\left(1-\frac{1}{2^{s}}\...

Show that:

$$\int_{0}^{\infty}\frac{(x+1)\ln^{2}(x+1)}{(4x^{2}+8x+5)^{3/2}}dx=\frac{\pi^{2}}{40}$$

I think I have another way using contours. $$\sin^{2}(u)=\frac{1-\cos(2u)}{2}$$ Now, consider $$f(z)=\frac{1-e^{2i\tan(z)}}{2z^{2}}$$ and use a rectangular contour with height y. There is a pole at z=0. sintanint.png [/centre] There are also what is called 'essential singularities' at $z=\pi k +\fra...

I used the residue at infinity and changed it to 0 by using $$Res(f(z), \infty)=-Res\left(\frac{1}{z^{2}}f(1/z), \;\ 0\right)$$ $$-2\pi i \cdot \lim_{z\to 0}\frac{1}{z}\cdot \frac{\prod_{k=0}^{5}(1-2k/z)}{\prod_{k=0}^{5}(1/z-2k)}$$ $$=-2\pi i \cdot 3840$$ One could also add up the residues of the po...