A "closed form expression" is any representation of a mathematical expression in terms of "known" functions, "known" usually being replaced with "elementary".

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263
votes
7answers
80k views

Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \ \mathrm dx$

I need help with this integral: $$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx.$$ The integrand graph looks like this: $\hspace{1in}$ The ...
100
votes
2answers
4k views

How to prove $\int_0^1\tan^{-1}\left[\frac{\tanh^{-1}x-\tan^{-1}x}{\pi+\tanh^{-1}x-\tan^{-1}x}\right]\frac{dx}{x}=\frac{\pi}{8}\ln\frac{\pi^2}{8}?$

How can one prove that $$\int_0^1 \tan^{-1}\left[\frac{\tanh^{-1}x-\tan^{-1}x}{\pi+\tanh^{-1}x-\tan^{-1}x}\right]\frac{dx}{x}$$ $$=\frac{\pi}{8}\ln\frac{\pi^2}{8}?$$
95
votes
3answers
23k views

Evaluate $\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx$

I am trying to find a closed form for $$\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx = 0.094561677526995723016 \cdots$$ It seems that the answer is $$\frac{\pi^2}{12}\left( 1-\...
83
votes
6answers
10k views

How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$

Please help me to find a closed form for this integral: $$I=\int_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx\tag1$$ I suspect it might exist because there are similar integrals having closed forms: $$\begin{...
80
votes
2answers
7k views

The Monster Integral

Compute the following integral \begin{equation} \int_0^{\Large\frac{\pi}{4}}\left[\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right] x\, \exp\left[\frac{x^2-1}{x^2+1}\right]\, ...
66
votes
4answers
6k views

An integral involving Airy functions $\int_0^\infty\frac{x^p}{\operatorname{Ai}^2 x + \operatorname{Bi}^2 x}\mathrm dx$

I need your help with this integral: $$\mathcal{K}(p)=\int_0^\infty\frac{x^p}{\operatorname{Ai}^2 x + \operatorname{Bi}^2 x}\mathrm dx,$$ where $\operatorname{Ai}$, $\operatorname{Bi}$ are Airy ...
65
votes
2answers
2k views

Conjecture $\int_0^1\frac{dx}{\sqrt[3]x\,\sqrt[6]{1-x}\,\sqrt{1-x\left(\sqrt{6}\sqrt{12+7\sqrt3}-3\sqrt3-6\right)^2}}=\frac\pi9(3+\sqrt2\sqrt[4]{27})$

Let $$\alpha=\sqrt{6}\ \sqrt{12+7\,\sqrt3}-3\,\sqrt3-6.\tag1$$ Note that $\alpha$ is the unique positive root of the polynomial equation $$\alpha^4+24\,\alpha^3+18\,\alpha^2-27=0.\tag2$$ Now consider ...
64
votes
1answer
1k views

Closed form for $\int_0^\infty\ln\frac{J_\mu(x)^2+Y_\mu(x)^2}{J_\nu(x)^2+Y_\nu(x)^2}\mathrm dx$

Consider the following integral: $$\mathcal{I}(\mu,\nu)=\int_0^\infty\ln\frac{J_\mu(x)^2+Y_\mu(x)^2}{J_\nu(x)^2+Y_\nu(x)^2}\mathrm dx,$$ where $J_\mu(x)$ is the Bessel function of the first kind: $$J_\...
62
votes
2answers
2k views

Integral $\int_1^\infty\frac{\operatorname{arccot}\left(1+\frac{2\pi}{\operatorname{arcoth}x-\operatorname{arccsc}x}\right)}{\sqrt{x^2-1}}\mathrm dx$

Consider the following integral: $$\mathcal{I}=\int_1^\infty\frac{\operatorname{arccot}\left(1+\frac{2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)}{\sqrt{x^2-1}}\mathrm dx,$$ where ...
60
votes
2answers
2k views

Conjecture $\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}\stackrel?=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi$

$$\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}\stackrel?=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi\tag1$$ The equality numerically holds up to at least $10^4$ decimal digits. ...
56
votes
2answers
668 views

Conjecture $_2F_1\left(\frac14,\frac34;\,\frac23;\,\frac13\right)=\frac1{\sqrt{\sqrt{\frac4{\sqrt{2-\sqrt[3]4}}+\sqrt[3]{4}+4}-\sqrt{2-\sqrt[3]4}-2}}$

Using a numerical search on my computer I discovered the following inequality: $$\left|\,{_2F_1}\left(\frac14,\frac34;\,\frac23;\,\frac13\right)-\rho\,\right|<10^{-20000},\tag1$$ where $\rho$ is ...
53
votes
2answers
2k views

Closed form for $\int_0^1\log\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\mathrm dx$

Please help me to find a closed form for the following integral: $$\int_0^1\log\left(\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\right)\,{\mathrm d}x.$$ I was told it could be calculated in a ...
48
votes
3answers
1k views

Infinite Series $\sum_{n=1}^\infty\frac{(H_n)^2}{n^3}$

How to prove that $$\sum_{n=1}^{\infty}\frac{(H_n)^2}{n^3}=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)$$ $H_n$ denotes the harmonic numbers.
46
votes
1answer
2k views

Is it possible to simplify $\frac{\Gamma\left(\frac{1}{10}\right)}{\Gamma\left(\frac{2}{15}\right)\ \Gamma\left(\frac{7}{15}\right)}$?

Is it possible to simplify this expression? $$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\ \Gamma\left(\frac{7}{15}\right)}$$ Is there a systematic ...
45
votes
4answers
4k views

Integral $\int_0^1\frac{\ln\left(x+\sqrt2\right)}{\sqrt{2-x}\,\sqrt{1-x}\,\sqrt{\vphantom{1}x}}\mathrm dx$

Is there a closed form for the integral $$\int_0^1\frac{\ln\left(x+\sqrt2\right)}{\sqrt{2-x}\,\sqrt{1-x}\,\sqrt{\vphantom{1}x}}\mathrm dx.$$ I do not have a strong reason to be sure it exists, but I ...
45
votes
1answer
2k views

The Wicked Integral

My brother's friend gave me the following wicked integral with a beautiful result \begin{equation} {\Large\int_0^\infty} \frac{dx}{\sqrt{x} \bigg[x^2+\left(1+2\sqrt{2}\right)x+1\bigg] \bigg[1-x+x^...
42
votes
3answers
4k views

A strange integral having to do with the sophomore's dream:

I recently noticed that this really weird equation actually carries a closed form! $$\int_0^1 \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x=0$$ I honestly do not know how to ...
42
votes
2answers
2k views

Integral $\int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \mathrm dx$

Regarding this problem, I conjectured that $$ I(r, s) = \int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \mathrm dx = 4 \pi \...
42
votes
1answer
1k views

Closed form for $\int_0^\infty\frac{\log\left(1+\frac{\pi^2}{4\,x}\right)}{e^{\sqrt{x}}-1}\mathrm dx$

I encountered this integral in my calculations: $$\int_0^\infty\frac{\log\left(1+\frac{\pi^2}{4\,x}\right)}{e^{\sqrt{x}}-1}\mathrm dx=2\int_0^\infty\frac{x\log\left(1+\frac{\pi^2}{4\,x^2}\right)}{e^x-...
40
votes
4answers
2k views

Integral $\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$

Is it possible to evaluate this integral in a closed form? $$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$$
40
votes
3answers
1k views

Prove $\int_{0}^{\pi/2}\frac{dx}{1+\sin^2{(\tan{x})}}=\frac{\pi}{2\sqrt{2}}\bigl(\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\bigr)$

Prove the following integral $$I=\int_{0}^{\frac{\pi}{2}}\dfrac{dx}{1+\sin^2{(\tan{x})}}=\dfrac{\pi}{2\sqrt{2}}\left(\dfrac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\right)$$ This integral result was ...
40
votes
5answers
867 views

Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$

Is there a closed form for the following infinite product? $$\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$$
40
votes
3answers
790 views

Conjectural closed-form representations of sums, products or integrals

What are some examples of infinite sums, products or definite integrals that have conjectural closed form representations that were confirmed by numerical calculations up to whatever maximum precision ...
39
votes
2answers
1k views

Evaluate $ \int_{0}^{\pi/2}\frac{1+\tanh x}{1+\tan x}dx $

I need the method to evaluate this integral (the closed-form if possible). $$ \int_{0}^{\pi/2}\frac{1+\tanh x}{1+\tan x}\,dx $$ I used the relationship between $\tan x$ and $\tanh x$ but it didn't ...
38
votes
3answers
818 views

Calculate the following infinite sum in a closed form $\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$?

Is it possible to calculate the following infinite sum in a closed form? If yes, please point me to the right direction. $$\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$$
38
votes
2answers
2k views

Integral $\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx$

Is it possible to evaluate this integral in a closed form? $$I=\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx$$ It also can be represented as $$I=\int_0^{\pi/4}\frac{\phi^2}{\cos \phi\,\sqrt{\cos 2\...
38
votes
1answer
947 views

Integral $\int_0^1\frac{x^9\left(x^4+x^2-x-1-5\ln x\right)}{\left(x^{10}-1\right)\ln x}\mathrm dx$

A friend of mine sent me an integral that she had not been able to crack, and me neither. It comes with a result, but without a proof (I suppose it originated in some math contest). Could you please ...
38
votes
1answer
1k views

Prove that $\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx =\frac{\pi e}{24} $

I've found here the following integral. $$I = \int_{0}^{1}\sin{(\pi (1-x))}x^x(1-x)^{1-x}\,dx=\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx=\frac{\pi e}{24}$$ I've never seen it before and I also didn'...
37
votes
4answers
2k views

Closed form for ${\large\int}_0^1\frac{\ln^2x}{\sqrt{1-x+x^2}}dx$

I want to find a closed form for this integral: $$I=\int_0^1\frac{\ln^2x}{\sqrt{x^2-x+1}}dx\tag1$$ Mathematica and Maple cannot evaluate it directly, and I was not able to find it in tables. A numeric ...
37
votes
3answers
1k views

An integral involving Fresnel integrals $\int_0^\infty \left(\left(2\ S(x)-1\right)^2+\left(2\ C(x)-1\right)^2\right)^2 x\ \mathrm dx,$

I need to calculate the following integral: $$\int_0^\infty \left(\left(2\ S(x)-1\right)^2+\left(2\ C(x)-1\right)^2\right)^2 x\ \mathrm dx,$$ where $$S(x)=\int_0^x\sin\frac{\pi z^2}{2}\mathrm dz,$$ $$...
37
votes
3answers
931 views

Find the value of $\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$

I need to find a closed-form for the following integral. Please give me some ideas how to approach it: $$\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$$
37
votes
1answer
663 views

How can I prove this closed form for $\sum_{n=1}^\infty\frac{(4n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^n}$

How can I prove the following conjectured identity? $$\mathcal{S}=\sum_{n=1}^\infty\frac{(4\,n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^n}\stackrel?=\frac{\sqrt3}{2\,...
36
votes
4answers
1k views

Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^32^n}$

I'm trying to find a closed form for the following sum $$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$ where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number. Could you help me with it?
35
votes
4answers
4k views

Do harmonic numbers have a “closed-form” expression?

One of the joys of high-school mathematics is summing a complicated series to get a “closed-form” expression. And of course many of us have tried summing the harmonic series $H_n =\sum \limits_{k \leq ...
35
votes
2answers
1k views

A conjectural closed form for $\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}$

Let $$S=\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!},\tag1$$ its numeric value is approximately $S \approx 0.517977853388534047...$${}^{[more\ digits]}$ $S$ can be represented in terms of the ...
35
votes
2answers
558 views

Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$

Please help me to find a closed form for the infinite product $$\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$$ where $\tanh(z)=\frac{e^z-e^{-z}}{e^z+e^{-z}}$ is the hyperbolic tangent.
35
votes
3answers
1k views

Prove $\displaystyle \int_{0}^{\pi/2} \ln \left(x^{2} + (\ln\cos x)^2 \right) \, dx=\pi\ln\ln2 $

How to prove $$ \int_{0}^{\pi/2}\ln\left(\,x^{2} + \ln^{2}\left(\,\cos\left(\,x\,\right)\,\right) \,\right)\,{\rm d}x\ =\ \pi\ln\left(\,\ln\left(\, 2\,\right)\,\right) $$ I don't know how to ...
34
votes
3answers
1k views

A conjectured closed form of $\int_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx$

Consider the following integral: $$\mathcal{I}=\int_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx.$$ I tried to evaluate $\mathcal{I}$ in a closed form (both manually and using ...
34
votes
6answers
1k views

Infinite Series $‎\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}$

‎If $f\left(z \right)=\sum_{n=2}^{\infty}a_{n}z^n$ and $\sum_{n=2}^{\infty}|a_n|$ converges then‎, $$\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\sum_{n=2}^{\infty}a_n\zeta\left(n\right)‎.$$ ‎Since ...
34
votes
3answers
1k views

Integral ${\large\int}_0^\infty\frac{\ln x}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}\ \mathrm dx$

Please help me to evaluate this integral: $$ I={\large\int}_{0}^{\infty}{\ln\left(x\right) \over 1 + x}\, \,\sqrt{\,x + \sqrt{\,1 + x^{2}\,}\, \over 1 + x^{2}\,}\,\,{\rm d}x.\tag1 $$ Mathematica could ...
34
votes
2answers
2k views

Integral $\int_0^{\pi/2}\arctan^2\left(\frac{6\sin x}{3+\cos 2x}\right)\mathrm dx$

Is it possible to evaluate this integral in a closed form? $$I=\int_0^{\pi/2}\arctan^2\left(\frac{6\sin x}{3+\cos 2x}\right)\mathrm dx$$
34
votes
3answers
2k views

Closed Form for $~\int_0^1\frac{\text{arctanh }x}{\tan\left(\frac\pi2~x\right)}~dx$

Does $$~\displaystyle{\int}_0^1\frac{\text{arctanh }x}{\tan\left(\dfrac\pi2~x\right)}~dx~\simeq~0.4883854771179872995286585433480\ldots~$$ possess a closed form expression ? This recent post, in ...
34
votes
2answers
621 views

Closed form for $\int_0^1\frac{x^{5/6}}{(1-x)^{1/6}\,(1+2\,x)^{4/3}}\log\left(\frac{1+2x}{x\,(1-x)}\right)\,dx$

I need to evaluate this integral: $$Q=\int_0^1\frac{x^{5/6}}{(1-x)^{1/6}\,(1+2\,x)^{4/3}}\log\left(\frac{1+2x}{x\,(1-x)}\right)\,dx.$$ I tried it in Mathematica, but it was not able to find a closed ...
33
votes
3answers
913 views

Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$

Here is a challenging one maybe some would like a go at. Show that: $$\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx=\frac{-9\pi^{4}}{256}+\frac{\pi^{3}}{32}\ln2+\frac{\pi^{2}}{6}G-\...
33
votes
6answers
1k views

Closed-form of $\int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx $

I'm looking for a closed form of this integral. $$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx ,$$ where $\operatorname{Li}_2$ is the dilogarithm function. A numerical ...
33
votes
1answer
684 views

Closed form for $\left(1+\left(\frac{1}{2}+\left(\frac{1}{3}+\left(\frac{1}{4}+\cdots\right)^2\right)^2\right)^2\right)^2$?

Nested squares seem to be more promising than nested radicals, since they give rational approximations and in principle can be expanded into a series. These two expressions converge numerically: $$\...
32
votes
4answers
1k views

Is $\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}$ true for $m\in\mathbb N$?

Question : Is the following true for any $m\in\mathbb N$? $$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$ Motivation : I reached $(\star)$ ...
32
votes
3answers
1k views

Closed form of $\displaystyle\mathscr{R}=\int_0^{\frac{\pi}{2}}\sin^2x\,\ln\big(\sin^2(\tan x)\big)\,\,dx$

Inspired by Mr. Olivier Oloa in this question. Does the following integral admit a closed form? \begin{align} \mathscr{R}=\int_0^{\Large\frac{\pi}{2}}\sin^2x\,\ln\big(\sin^2(\tan x)\big)\,\,dx \...
32
votes
3answers
763 views

A closed form for $\int_0^1{_2F_1}\left(-\frac{1}{4},\frac{5}{4};\,1;\,\frac{x}{2}\right)^2dx$

Is it possible to evaluate in a closed form integrals containing a squared hypergeometric function, like in this example? $$\begin{align}S&=\int_0^1{_2F_1}\left(-\frac{1}{4},\frac{5}{4};\,1;\,\...
31
votes
3answers
1k views

How can I prove $\pi=e^{3/2}\prod_{n=2}^{\infty}e\left(1-\frac{1}{n^2}\right)^{n^2}$?

I am interested about some infinite product representations of $\pi$ and $e$ like this. Last week I found this formula on internet $$\pi=e^{3/2}\prod_{n=2}^{\infty}e\left(1-\frac{1}{n^2}\right)^...