Recent content by sbhatnagar

I am sorry.:D I thought it would be nice to give some ideas on Euler Sums first. Anyway, if it is against the rules I shall not post a solution so early.

I think these kind of problems have gone out of fashion these days. Anyway, here is my solution to (1). Problem 1 Step 1 - Reduction to Euler Sum \[ \begin{aligned} \int_0^1 \frac{\log(1+x)\log(x)}{1-x}dx &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\int_0^1 \frac{x^n \log(x)}{1-x}dx \\ &=...

Thank you chisigma for that nice paper. :D My solution was different from the one given in it. I used Dilogarithms to evaluate it.

Prove that \[\int_0^1 \frac{\log(1+x)\log(x)}{1-x}dx=\zeta(3)-\frac{\pi^2}{4}\log(2)\] \[\int_0^1 \frac{\log(1+x^2)}{1+x}dx=\frac{3}{4}\log^2(2) -\frac{\pi^2}{48}\]

Hi everyone ;) I have a challenging problem which I would like to share with you. Prove that \[\frac{1}{2^2}+ \frac{1}{3^2} \left(1+\frac{1}{2} \right)^2+\frac{1}{4^2} \left( 1+\frac{1}{2} +\frac{1}{3}\right)^2 + \frac{1}{5^2} \left( 1+\frac{1}{2} +\frac{1}{3}+\frac{1}{4}\right)^2 +\cdots=...

Hi! :D I think I know how to solve this integral but my method is a little tedious. Anyway I will post my solution. Let $\displaystyle I(n) = \int_0^{\pi/2}\sin ^n (x)dx$ $I(n)$ can be evaluated in terms of gamma function. $$ I(n)= \int_0^{\pi/2}\sin ^n (x)dx = \frac{\sqrt{\pi} \Gamma \left(...

The integral may be rewritten as $$ \begin{aligned} I=\int_0^\infty e^{-x^2}\cos(kx) dx &= \frac{1}{2}\int_{-\infty}^{\infty} e^{-x^2}\cos(kx) dx \\ &= \frac{1}{2} \text{Re} \left[\int_{-\infty}^{\infty} e^{-x^2+ikx} dx\right]\end{aligned}$$ Here, we can use the general formula $$...

Hi Lord!:D My solution requires the following three facts: $$ \log\tan^{2}\theta = -4 \sum_{n \ \mathrm{odd}}^{\infty}\frac{1}{n}\cos 2n\theta $$ $$ \sum_{n \ \mathrm{odd}}^{\infty} \frac{1}{n} x^{n} = \frac{1}{2}\log\left(\frac{1+x}{1-x}\right) $$ and $$ \int_{0}^{\infty} \frac{t \sin...

Here's another method to do it without using differentiation under the integral sign. $$ \begin{aligned} I &= \int_0^1 \frac{t^2-1}{(t^2+1) \ln(t)}dt \\ &= \int_0^1 \frac{t+1}{t^2+1}\frac{t-1}{\ln(t)}dt \\ &= \int_0^1 \frac{t+1}{t^2+1} \int_0^1 t^x dx \ dt \\ &= \int_0^1 \int_0^1...

This is classic one. Prove that $$ \int_0^\infty \frac{dx}{\left\{x^4+(1+2\sqrt{2})x^2+1 \right\}\left\{x^{100}-x^{99}+x^{98}-\cdots +1\right\}}=\frac{\pi}{2(1+\sqrt{2})}$$

Cool Problem!(Drunk) I will solve this using differentiation under the integral sign. The integral can be written in another form $$ \int_0^\infty \frac{\tanh(x)}{x e^x}dx = \int_0^1 \frac{t^2-1}{(t^2+1)\ln t}dt$$ Let us define $$ I(\alpha) = \int_0^1 \frac{t^\alpha-1}{(t^2+1)\ln t}dt$$...

Very Good ZaidAlyafey! :)

Bacterius, my approach was a little different from yours. After using the substitution \(u=\ln(\tan x)\), I did \[\begin{align} \int_0^\infty \cos(x) \text{sech}(x)dx &= 2\int_0^\infty \frac{e^{-x}}{1+e^{-2x}}\cos(x) dx\end{align}\] Now use \(\displaystyle...

Prove that \[\int_0^{\pi/2}\cos(\ln(\tan(x)))dx=\frac{\pi}{2} \text{sech} \left(\frac{\pi}{2}\right)\]

I will post the solution now. In the integral, \[ I=\int_0^1 x^{a-1}(1-x^c)^{b-1}\ln x \ dx\] substitute \(t=x^c\), and obtain \[ I=\frac{1}{c^2}\int_0^1 t^{\frac{a}{c}-1}(1-t)^{b-1}\ln t \ dt\] This integral can be evaluated using the result obtained in my previous post...