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I am gonna delete my post. It's too cliche'. I want to do it differently. That is cool. Leave to ol' Seraphim:) When n=3, doesn't it lead to something maybe easier to deal with: $$\int_{0}^{\frac{\pi}{2}}\ln^{3}(2\cos(x))=\int_{0}^{\frac{\pi}{2}}3x^{2}\ln(2\cos(x))$$ That's a clever approach, T. EDI...

deleted.

Wow, cool beans T.

It's fun discovering other things while researching another.

That is where I have found some cool stuff I submit.

If we use parts on this critter by letting $u=\tan^{-1}(x), \;\ dv=\frac{x^{a-1}}{(1+x^{a})^{2}}dx$, then the resulting integral is the famous: $$\frac{1}{|a|}\int_{0}^{\infty}\frac{1}{(x^{2}+1)(x^{a}+1)}dx$$ This integral resolves to $\frac{\pi}{4}$ regardless of the value of 'a'. It is related to ...

This one isn't super nasty if we use the sub $x=\tan^{-1}(t)$

I did not finish this one. I just noticed that the mentioned sub leads to:

$$4\int_{0}^{\infty}\left(\frac{\ln(x)}{x-1}\right)^{4}dx$$

This looks more manageable....I think.

Hi Gang. Those last few posted were little challenge to you all. Cool though.

Here is one that has a little more meat on it. Appears to anyway.


$$\int_{0}^{\pi/2}\left(\frac{\log(\tan(x))}{\sin(x-\frac{\pi}{4})}\right)^{4}\cos^{2}(x)dx=32\zeta(2)+64\zeta(4)$$

Show that:

$$\int_{0}^{\frac{\pi}{2}}\log\left(\frac{1+\cos^{2}\theta}{\sqrt{1+1/8\cos^{2}\theta}}\right)d\theta = \frac{\pi}{4}\ln(2)$$

I see no one has posted a reply to this one yet. I managed to get it by using 'parts'. First parts: Let $$u=\frac{\sin(x)}{x+1}, \;\ du=\frac{(x+1)\cos(x)-\sin(x)}{(x+1)^{2}}dx, \;\ dv=\frac{e^{-x}(x^{2}+3x+3)}{(x+1)^{2}}dx, \;\ v=\frac{-e^{-x}(x+2)}{x+1}$$ $$\frac{-e^{-x}(x+2)\sin(x)}{(x+1)^{2}}+\i...

Show that:

$$\int_{0}^{\infty}\left(\frac{\sin(x)}{x}\right)^{2}e^{-2ax}dx=a\log\left(\frac{a}{\sqrt{1+a^{2}}}\right)+\cot^{-1}(a)$$

By using the series for -ln(1-x), we can establish the equivalency of $$-\sum_{n=1}^{\infty}\frac{e^{ina}}{n}x^{n-1}=\frac{\ln|1-xe^{ia}|}{x}=\frac{\ln(x^{2}-2x\cos(a)+1)}{2x}$$ Take the real part and we find that: $$-\sum_{n=1}^{\infty}\frac{\cos(na)}{n}x^{n-1}=\frac{\ln|1-xe^{ia}|}{x}=\frac{\ln(x^...