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2 votes
1 answer
177 views

Show that $ \int_0^{\pi\over 2}\frac{\sin(2nx)}{\sin^{2n+2}(x)}\frac{1}{e^{2\pi \cot x}-1}dx =(-1)^{n-1}\frac{2n-1}{4(2n+1)} $

Show that $$ \int_0^{\pi\over 2}\frac{\sin(2nx)}{\sin^{2n+2}(x)}\frac{1}{e^{2\pi \cot x}-1}dx =(-1)^{n-1}\frac{2n-1}{4(2n+1)} $$ My attempt Lemma-1 \begin{align*} \frac{\sin(2nx)}{\sin^{2n}(x)}&=\...
Mods And Staff Are Not Fair's user avatar
0 votes
0 answers
132 views

Calculation of $\sum_{n=1}^{\infty}\frac{\psi_1(n)}{2^nn^2}$

Calculation of $$\sum_{n=1}^{\infty}\frac{\psi_1(n)}{2^nn^2}$$ My attempt \begin{align*} \sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n^2} &= -\sum_{n=1}^\infty\psi_1(n)\left(\frac{\log(2)}{2^n n}+\int_0^...
Mods And Staff Are Not Fair's user avatar
8 votes
2 answers
244 views

How to calculate $\int _0^1 \int _0^1\left(\frac{1}{1-xy} \ln (1-x)\ln (1-y)\right) \,dxdy$

Let us calculate the sum $$ \displaystyle{\sum_{n=1}^{+\infty}\left(\frac{H_{n}}{n}\right)^2}, $$ where $\displaystyle{H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}}$ the $n$-th harmonic number. My try The ...
Mods And Staff Are Not Fair's user avatar
1 vote
1 answer
95 views

Evaluation of $\int_{0}^{\frac{\pi}{4}} \frac{\log(\log(\tan(\frac{\pi}{4} + x))) \cdot \log(\tan(\frac{\pi}{4} + x)))}{\tan(2x)} \,dx$ [closed]

$$\int_{0}^{\frac{\pi}{4}} \frac{\log(\log(\tan(\frac{\pi}{4} + x))) \cdot \log(\tan(\frac{\pi}{4} + x)))}{\tan(2x)} \,dx$$ $$\int_{0}^{\frac{\pi}{4}} \frac{\log(\log(\tan(\frac{\pi}{4} + x))) \cdot \...
Mods And Staff Are Not Fair's user avatar
5 votes
2 answers
157 views

Show that $\sum_{n=1}^{\infty} \frac{\binom{2n}{n} (H_{2n} - H_n)}{4^n (2n - 1)^2} = 2 + \frac{3\pi}{2} \log(2) - 2G - \pi$

Show that $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n} (H_{2n} - H_n)}{4^n (2n - 1)^2} = 2 + \frac{3\pi}{2} \log(2) - 2G - \pi$$ My try : We know that $$\sum_{n=1}^{\infty} \binom{2n}{n} (H_{2n} - H_{n}) ...
Mods And Staff Are Not Fair's user avatar
3 votes
2 answers
206 views

why does $\pi$ always show up in $\int_0 ^1 \frac{x^c}{1+x^k} dx$ if $c\neq mk-1$ for all $m \in \mathbb{N}$

when I posted this question I was interested in the sum $ \sum_{n=0} ^{\infty} \frac{(-1)^n}{4n+3}$ but when I thought about the generalised sum $ \sum_{n=0} ^{\infty} \frac{(-1)^n}{kn+c +1}$ for all $...
pie's user avatar
  • 6,620
11 votes
3 answers
451 views

How to evaluate $ \sum\limits_{k=0} ^{\infty} \frac{(-1)^k}{4k+3}$?

I was trying to solve the integral $\int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}}dx$ and I noticed I can do the following: $$\int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}}dx=\int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}} \...
pie's user avatar
  • 6,620
5 votes
2 answers
185 views

Generating Function $\sum_{k=1}^{\infty}\binom{2k}{k}^{-2}x^{k}$

Closed Form For : $$S=\sum_{k=1}^{\infty}\binom{2k}{k}^{-2}x^{k}$$ Using the Series Expansion for $\arcsin^2(x)$ one can arrive at : $$\sum_{k=0}^{\infty}\binom{2k}{k}^{-1}x^{k}=\frac{4}{4-x}-4\arcsin\...
Miracle Invoker's user avatar
1 vote
0 answers
60 views

Integration including the floor function

I'm aware of the way to quantify the following integral in the following way, however I'm trying to find another way to express the given integral. Especially when the function $f(x)$ is not ...
Mallophas's user avatar
2 votes
3 answers
124 views

Evaluating $\int_{0}^{1}{\tan^{-1}(x)\, dx}$ via a series (not IBP)

I tried : $$\begin{align}I&=\int_{0}^{1}{\tan^{-1}(x)\, dx}\\&=\int_{0}^{1}{\sum_{k\geq0}{\frac{(-1)^kx^{2k+1}}{2k+1}}\, dx}\\&=\sum_{k\geq0}{\frac{(-1)^k}{2k+1}\left[\frac{x^{2(k+1)}}{2(k+...
AnthonyML's user avatar
  • 977
4 votes
0 answers
135 views

Simplify a summation in the solution of $\displaystyle\int_{0}^{\infty}e^{-cx}x^{n}\arctan(ax)\mathrm{d}x$

Context I calculated this integral: $$\begin{array}{l} \displaystyle\int_{0}^{\infty}e^{-cx}x^{n}\arctan(ax)\mathrm{d}x=\\ \displaystyle\frac{n!}{c^{n+1}}\left\lbrace\sum_{k=0}^{n}\left[\text{Ci}\left(...
Math Attack's user avatar
3 votes
1 answer
129 views

Compact form of solution of $\displaystyle\int_0^{\frac{\pi}{2}}\ln\left(1+\alpha^n\sin(x)^{2n}\right)\mathrm{d}x$

I hope it won't be categorized as a trivial question, I solved this integral and arrived at the following form: $${\int_{0}^{\frac{\pi}{2}}\ln\left(1+\alpha^n\sin(x)^{2n}\right)\mathrm{d}x=\frac{n\pi}{...
Math Attack's user avatar
1 vote
1 answer
153 views

Sum of values across a line?

I've been thinking recently and have confused myself a little bit as to the difference between sums and integrals. I understand that an integral is a Riemann sum where you take the limit as the ...
Dr-Galunga's user avatar
7 votes
3 answers
292 views

Solve $\int_0^\infty\frac x{e^x-e^{\frac x2}}dx$

I was able to solve the integral $$\int_0^\infty\frac x{e^x-e^\frac x2}dx=4\left(\frac{\pi^2}6-1\right)$$ I want to see other approaches to solving it. Here is my solution: $$\int_0^\infty\frac x{e^x-...
Kamal Saleh's user avatar
  • 6,549
7 votes
3 answers
277 views

Evaluating a Logarithmic Integral

For everything on this post $n$ and $m$ are positive integers. The other day I found the following integral on the popular post "Integral Milking" and decided to give it a go. $$\large\int_{...
Alejandro Jimenez Tellado's user avatar

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