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70 questions with no upvoted or accepted answers
16 votes
0 answers
818 views

Juantheron-like integral

While seeing this post, the following integral is just struck me \begin{equation} \int_0^\infty \frac{dx}{(1+x^2)(1+\tan x)}\tag1 \end{equation} I have tried like what user @OlivierOloa did in his ...
Sophie Agnesi's user avatar
10 votes
0 answers
262 views

Integrals involving powers and beta (or hypergeometric) function

I have the three following integrals, very similar the one to the others, $$I_1^{(p)}(N)\equiv\frac{1}{2^{N+p}}\int_0^1(1+t)^{N-1}(1-t)^pB\left(\frac{1}{t+1};N+p+1,N\right)\text{d}t$$ $$I_2^{(p)}(...
ARWarrior's user avatar
  • 299
10 votes
0 answers
2k views

Exact values of error function

The error function is defined as $$\operatorname{erf}(z)=\frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} \, dt.$$ We know that the Gaussian integral is $$\int_{-\infty}^{\infty} e^{-x^2}\,dx=\sqrt{\pi}.$$ ...
user153012's user avatar
  • 12.4k
9 votes
0 answers
444 views

Why do these two integrals use roots of reciprocal polynomials?

There is the nice integral by V. Reshetnikov, $$\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{3}\;\alpha}\tag1$$ also discussed in this post. By ...
Tito Piezas III's user avatar
9 votes
0 answers
387 views

Closed form of $\sum _{n=0}^{\infty} \frac{\left(-\pi ^2\right)^n \cos \left(2^nb\right)}{(2 n)!}$

Is it possible to calculate the sum $$ \sum _{n=0}^{\infty} \frac{\left(-\pi ^2\right)^n \cos \left(2^nb\right)}{(2 n)!} $$ in closed form? Formal naive argument gives $$ \sum _{n=0}^{\infty} \...
Tyrell's user avatar
  • 1,695
8 votes
0 answers
413 views

More on the log sine integral $\int_0^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\mathrm{d}\theta$

I. In this post, the OP asks about the particular log sine integral, $$\mathrm{Ls}_{7}^{\left ( 3 \right )} =-\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\,\mathrm{d}\...
Tito Piezas III's user avatar
8 votes
0 answers
143 views

More on $\sum_{n=1}^\infty\frac{(4n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^n}$

Let, $$\alpha=2\sqrt[3]{1+\sqrt2}-\frac2{\sqrt[3]{1+\sqrt2}}$$ In this post, it was asked if, $$\sum_{n=1}^\infty\frac{(4\,n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^...
Tito Piezas III's user avatar
8 votes
0 answers
415 views

Improper Integral $\int_0^\infty\tan\left(\frac x{\sqrt{x^3+x^2}}\right)\frac{\ln(1+\sqrt x)}xdx$

This integral is from integral Find $$\int_0^\infty\tan\left(\frac x{\sqrt{x^3+x^2}}\right)\frac{\ln(1+\sqrt x)}xdx$$ I have get $$\int_0^\infty\tan\left(\frac x{\sqrt{x^3+x^2}}\right)\frac{\ln(1+\...
math110's user avatar
  • 93.6k
7 votes
1 answer
295 views

how to evaluate $\int_0^{\infty} \frac{x \ln ^2\left(1-e^{-2 \pi x}\right)}{e^{\frac{\pi x}{2}}+1} d x$

Question: how to evaluate $$\int_0^{\infty} \frac{x \ln ^2\left(1-e^{-2 \pi x}\right)}{e^{\frac{\pi x}{2}}+1} d x$$ MY try to evaluate the integral $$ \begin{aligned} & I=\int_0^{\infty} \frac{x \...
user avatar
7 votes
0 answers
276 views

Evaluate $\int_{0}^{1} \frac{K(k)E(k)^2-\frac{\pi^3}{8} }{k} \text{d}k$ and $\int_{0}^{1} \frac{E(k)^3-\frac{\pi^3}{8} }{k} \text{d}k$

Let $K(k),E(k)$ be the complete elliptic integral of the first kind and second kind respectively, where $k$ is the elliptic modulus. Consider four integrals, $$\begin{aligned} &I_1=\int_{0}^{1} \...
Setness Ramesory's user avatar
7 votes
0 answers
302 views

Closed-form of integrals containing double exponentials

Are there closed forms for the following integrals? $$\begin{align} I_1(w) & = \int_{-\infty}^{\infty} \frac{\exp(-we^y)}{y^2+\pi^2} dy, \\ I_2(w) & = \int_{-\infty}^{\infty} \frac{\exp(-we^y+...
user153012's user avatar
  • 12.4k
6 votes
0 answers
104 views

Please help me identify any errors in my solution to the following DE: $xf(x)-f'(x)=0$, $f(0)=1$

Context/background: I am self-studying series, first in the context of generating functions and now in the context of functional/differential equations. As such, I like to set myself practise problems,...
H. sapiens rex's user avatar
6 votes
0 answers
172 views

How to evaluate $\int_0^1 \dfrac{\operatorname{Li}_2\left(\frac{x}{4}\right)}{4-x}\,\log\left(\dfrac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\right)\,dx$

crossposted: https://mathoverflow.net/q/464839 How to evaluate $$\int_0^1 \dfrac{\operatorname{Li}_2\left(\frac{x}{4}\right)}{4-x}\,\log\left(\dfrac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\right)\,dx=\dfrac{\pi^...
Mods And Staff Are Not Fair's user avatar
6 votes
0 answers
270 views

Integral of a product of five Bessel functions of order $0$

Does the following integral have a closed form? $$ \mathcal{J}(2,3,5,7,11) = \int_0^\infty x J_0(x\sqrt{2})J_0(x\sqrt{3})J_0(x\sqrt{5})J_0(x\sqrt{7})J_0(x\sqrt{11})\,dx. $$ I know that some similar ...
Kirill's user avatar
  • 14.6k
5 votes
0 answers
326 views

Converting $ \int_0^{\infty} \frac{e^{-\varepsilon s} \, (s+s^2)^{\beta}}{\log^{\gamma}((1+s)/s)} \, ds$ to a sum?

Could anyone shed some light on how to convert the following integral to a sum? $$ I=\int_0^{\infty} \frac{e^{-\varepsilon s} \, (s+s^2)^{\beta}}{\log^{\gamma}((1+s)/s)} \, ds; \qquad\,\,\varepsilon,\...
Z. Alfata's user avatar
  • 1,691

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