All Questions
81
questions
32
votes
1
answer
818
views
On the relationship between $\Re\operatorname{Li}_n(1+i)$ and $\operatorname{Li}_n(1/2)$ when $n\ge5$
Motivation
$\newcommand{Li}{\operatorname{Li}}$
It is already known that:
$$\Re\Li_2(1+i)=\frac{\pi^2}{16}$$
$$\Re\Li_3(1+i)=\frac{\pi^2\ln2}{32}+\frac{35}{64}\zeta(3)$$
And by this question, ...
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:
$$\int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}}dx=\int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}} \...
10
votes
5
answers
631
views
Evaluate $\int_{0}^{\pi} \frac{x\coth x-1}{x^2}dx$
I've been trying to evaluate certain series recently, and I found that
$$\sum_{r=1}^{\infty}\frac{1}{r}\arctan\frac{1}{r}=\frac{\pi}{2}\int_{0}^{\pi} \frac{x\coth x-1}{x^2} \, dx$$
Therefore, I would ...
9
votes
3
answers
2k
views
Evaluating $\int_0^1 \frac{\ln^m (1+x)\ln^n x}{x}\; dx$ for $m,n\in\mathbb{N}$
Evaluate $\displaystyle \int\limits_0^1 \dfrac{\ln^m (1+x)\ln^n x}{x}\; dx$ for $m,n\in\mathbb{N}$
I was wondering if the above had some kind of a closed form, here some of the special cases have ...
8
votes
2
answers
244
views
How to calculate $\int _0^1 \int _0^1\left(\frac{1}{1-xy} \ln (1-x)\ln (1-y)\right) \,dxdy$
Let us calculate the sum
$$
\displaystyle{\sum_{n=1}^{+\infty}\left(\frac{H_{n}}{n}\right)^2},
$$
where $\displaystyle{H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}}$ the $n$-th harmonic number.
My try
The ...
8
votes
3
answers
407
views
An integration-via-summation formula
For symbolic transformation of integrals and series I occasionally use this formula:
$$\int_0^1f(x)\,dx=-\sum_{n=1}^\infty\sum_{m=1}^{2^n-1}\frac{(-1)^m}{2^n}f\left(\frac m{2^n}\right)\tag{$\diamond$}$...
8
votes
1
answer
251
views
On the integral $\int_1^\infty\big(\{x\}^n-\frac1{n+1}\big)\frac{dx}x$
According to Dirichlet's test (integral version),
$$
I_n=\int_1^\infty\big(\{x\}^n-\frac1{n+1}\big)\frac{dx}x
$$
converges, where $n$ is a positive integer and $\{x\}$ denotes the fractional part of $...
8
votes
1
answer
317
views
Evaluating the integral $\frac{1}{2^{2n-2}}\int_0^1\frac{x^{4n}\left(1-x\right)^{4n}}{1+x^2} dx$
Prove that :
$$ \frac{1}{2^{2n-2}}\int \limits_{0}^{1} \dfrac{x^{4n}\left(1-x\right)^{4n}}{1+x^2} dx =$$$$\sum \limits_{j=0}^{2n-1}\dfrac{(-1)^j}{2^{2n-j-2}\left(8n-j-1\right)\binom{8n-j-2}{4n+j}} + (-...
7
votes
3
answers
292
views
Solve $\int_0^\infty\frac x{e^x-e^{\frac x2}}dx$
I was able to solve the integral $$\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-...
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.
$$\large\int_{...
7
votes
0
answers
453
views
Can we interchange the Integral and Summation when a limit is $\infty$?
I was trying to Evaluate the Integral:
$$\Large{I=\int_1^{\infty} \frac{\ln x}{x^2+1} dx}$$
$$\color{#66f}{{\frac{1}{x^2+1} = \frac{1}{x^2\left(1+\frac{1}{x^2}\right)}=\frac{1}{x^2}\cdot \frac{1}{1+...
6
votes
1
answer
353
views
Deriving the Integral for Alternating Harmonic Series Partial Sums
The partial sums of the harmonic series (the Harmonic Number, $H_n$) are given by
$$H_n=\sum_{k=1}^{n} \frac{1}{k}$$
and the well known integral representation is
$$H_n=\int_0^1 \frac{1-x^n}{1-x}\,dx$$...
6
votes
0
answers
535
views
Integral $I=\int_0^1 \frac{\log x \log (1+x) \log(1-x) \log(1+x^2)\log(1-x^2)}{x^{3/2}}dx$
Hi I am trying to integrate and obtain a closed form result for
$$
I:=\int_0^1 \frac{\log x \log (1+x) \log(1-x) \log(1+x^2)\log(1-x^2)}{x^{3/2}}dx.
$$
Here is what I tried (but I do not think this is ...
5
votes
3
answers
294
views
Find the closed-form solution to a integral with the floor function
I have the following integral:
$$\int_0^1k^{\left\lfloor\frac{1}{x}\right\rfloor}dx$$
My question is, is there a nice closed form of this integral?
I have not any idea where to start.
Maybe I could ...
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\...