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That was a fun one. Kudoes fellas

Thanks for finishing, rd

Hey T. Heck, I think this one is more involved than that triple one I managed to get sometime back. Anyway, I think I have it. Though, I may have to come back to finish posting. $$\int_{0}^{1}x^{n}\left\{\frac{1}{x}\right\}^{n}dx$$ Let $x=1/u, \;\ dx=\frac{-1}{u^{2}}du$ $$\begin{align*}\int_{1}^{\in...

Well, since no one bit, I reckon I will post a little something on the log integral. $$\int_{0}^{\frac{\pi}{2}}\log(a^{2}\sin^{2}(x)+b^{2}\cos^{2}(x))dx$$ Let $u=\tan(x), \;\ dx=\frac{1}{u^{2}+1}du$ $$\int_{0}^{\infty}\frac{\log\left(\frac{a^{2}u^{2}+b^{2}}{u^{2}+1}\right)}{u^{2}+1}du$$ $$=\int_{0}^...

Hi T

No, I do not have a relation between the two. I just noticed they were the same answer and thought maybe it would be cool to show.

Yes, I saw RD's solution to the other similar one. Clever.

Sorry for the typo on the first integral. I do not know why I had an 'n' there. But, maybe we can generalize. I reckon these two are not interesting or too easy?. But, for the case when n=1, a simple unit circle contour will suffice quite easily. Use the usual $$\cos(x)=\frac{z+z^{-1}}{2}, \;\ \sin(...

I was kind of waiting for someone to post, but I will go ahead. Simply note that the geometric series $$1+x+x^{2}+x^{3}+x^{4}+....=\sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}$$ is the same as the factorization: $$(1+x)(1+x^{2})(1+x^{4})(1+x^{8})\cdot\cdot\cdot (1+x^{2^{n}})=\prod_{n=0}^{\infty}(1+x^{2^{n...

Show that:

$$\int_{0}^{2\pi}\frac{(1-4\sin^{2}(x))\cos(2x)}{2-\cos(x)}dx=\frac{2\pi (91-52\sqrt{3})}{\sqrt{3}}$$


and

Evaluate:

$$\int_{0}^{\frac{\pi}{2}}\log\left(a^{2}\sin^{2}(x)+b^{2}\cos^{2}(x)\right)dx, \;\ 0<a<b$$

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\left(\left\{\frac{x}{y}\right\}\left\{\frac{y}{z}\right\}\left\{\frac{z}{x}\right\}\right)^{2}dxdydz$$ $$=\int_{0}^{1}\int_{0}^{1}\left(\left\{\frac{x}{y}\right\}\left\{\frac{1}{x}\right\}\right)^{2}dxdy$$ Break up integral at y: $$\int_{0}^{x}\int_{0}^{1}\left...

check this one out: http://integralsandseries.prophpbb.com/post3814.html#p3814 I am pretty sure the curly brackets represent fractional part, thought the OP should have stated that. I have seen the double integral version, so I am pretty sure a general form can be derived for this. What I mean is so...