Recent content by Shobhit

Write the integrand as $$\frac{1+x^2 \log(x)}{x+x^2 \log(x)}= 1+\frac{1}{x}-\frac{1+\log(x)}{1+x\log(x)}$$ Now integrate: $$ \int \frac{1+x^2 \log(x)}{x+x^2 \log(x)}dx=x+\log(x)-\log(1+x\log(x))+C$$

Over the years I have gained a lot of experience in evaluating definite integrals. So that's why I knew which trick can be applied here. The reason behind that particular definition of $T_n$ is the symmetry of the integral.

The integral at the end of your post is actually $T_{n-1}$. I know another solution using the residue theorem but I think that would be too hard for you to understand and probably not appropriate for the calculus section. :)

Let $$T_n=\int_0^{\pi/2}\frac{\sin((2n+1)x)}{\sin x}dx$$ Then $$ \begin{align*} T_n-T_{n-1} &= \int_0^{\pi/2} \frac{\sin((2n+1)x)-\sin((2n-1)x)}{\sin x}dx \\ &= 2 \int_0^{\pi/2}\cos(2nx)\; dx \\ &= \frac{1}{n}\sin(n\pi) \\ &= 0 \end{align*} $$ So $$T_n=T_{n-1}=T_{n-2}=\cdots =T_0 =...

Show that \[\prod_{k=2}^{\infty} \left(\frac{2k+1}{2k-1}\right)^{k} \left(1-\frac{1}{k^2}\right)^{k^2}=\frac{\sqrt{2}}{6}\pi \] This problem can be solved using only elementary methods. :D

Yes, that's a good paper. By the way, I worked out the case $x=\frac{1}{3}$: $$\sum_{n=1}^\infty \frac{\log(n)}{n^2}\cos\left(\frac{2\pi}{3}n \right)=\frac{\pi^2}{36}\log\left(\frac{12\pi^2 e^{2\gamma}}{A^{24}} \right)$$

The case $x=\frac{1}{4}$ gives $$\sum_{n=1}^\infty \frac{\log(n)}{n^2}\cos \left( \frac{\pi}{2}n\right)=\frac{\pi^2}{48}\log \left(\frac{2\pi e^{\gamma}}{A^{12}} \right)$$

I got $$ \sum_{n=1}^\infty \frac{\log(n)\cos(2\pi n x)}{n^2}=-\zeta'(2)+\pi^2 \left(x-x^2\right)\left( \log(2\pi)+\gamma -1\right) -2\pi^2 x \log \Gamma(x)+2\pi^2 \log \text{G}(x+1)+\frac{\pi}{2}\text{Cl}_2(2\pi x) $$ where $G(z)$ denotes the Barnes G Function and $\text{Cl}_2(z)$ is the...

Yes, that is how they can be solved. You may have to use equations (3) and (6) on this page.

The Riemann hypothesis (in the language of Integrals and Series ): $$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt=\frac{\pi(3-\gamma)}{32}$$ where $\gamma$ denotes the Euler Mascheroni Constant. Reference: V. V. Volchkov, On an equality...

Show that $$ \begin{align*} \sum_{n=0}^\infty \binom{2n}{n}^3 \frac{(-1)^n}{4^{3n}} &= \frac{\Gamma\left(\frac{1}{8}\right)^2\Gamma\left(\frac{3}{8}\right)^2}{2^{7/2}\pi^3} \tag{1}\\ \sum_{n=0}^\infty \binom{2n}{n}^3 \frac{1}{4^{3n}}&= \frac{\pi}{\Gamma \left(\frac{3}{4}\right)^4}\tag{2}...

Well done RV! :) Now, I am going to modify this problem slightly to make it even more challenging.Show that $$ \int_0^{\pi\over 2}\frac{\log(\tan x)}{\sqrt{2} \sin(x)+\sqrt{1+\sin^2 x}}dx = \frac{1}{\sqrt{2\,\pi}}\left(1+\frac{\log 2}{4}...

Show that $$\int_0^{\frac{\pi}{2}}\frac{\log \tan \theta}{\sqrt{1+\cos^2 \theta}}d\theta = \frac{\log 2}{16 \Gamma \left(\frac{3}{4} \right)^2}\sqrt{2\pi^3}$$ This integral is harder than the http://mathhelpboards.com/challenge-questions-puzzles-28/integration-challenge-7720.html. :D