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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224
votes
7answers
74k 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 ...
95
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 $$\int_0^1 \tan^{-1}\left[\frac{\tanh^{-1}x-\tan^{-1}x}{\pi+\tanh^{-1}x-\tan^{-1}x}\right]\frac{dx}{x}$$ $$=\frac{\pi}{8}\ln\frac{\pi^2}{8}?$$
76
votes
6answers
8k 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: ...
71
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\, ...
68
votes
3answers
5k 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( ...
63
votes
4answers
5k 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 ...
62
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: ...
60
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 ...
59
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 ...
53
votes
1answer
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. ...
53
votes
2answers
643 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 ...
46
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 ...
44
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 ...
42
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 ...
42
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] ...
39
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? $$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$$
39
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 ...
38
votes
5answers
818 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? $$\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$$
38
votes
3answers
765 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 ...
38
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 ...
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
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 ...
37
votes
3answers
770 views

$\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$

Is it possible to calculate the following infinite sum in a closed form? If yes, please point me to the right direction. $$\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$$
37
votes
1answer
923 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 ...
37
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 ...
36
votes
3answers
883 views

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

I need to find a closed-form for the following integral. Please give me some ideas how to approach it: $$\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$$
36
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 ...
35
votes
2answers
531 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
1answer
977 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 ...
34
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 ...
33
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 ...
33
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 ...
33
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 ...
33
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 ...
32
votes
4answers
1k views

A sum containing harmonic numbers $\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$

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?
32
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? $$I=\int_0^{\pi/2}\arctan^2\left(\frac{6\sin x}{3+\cos 2x}\right)\mathrm dx$$
32
votes
3answers
723 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? ...
32
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 ...
32
votes
2answers
591 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 ...
31
votes
4answers
949 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 ...
31
votes
2answers
2k views

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

This is somewhat similar to my previous question: Closed form for $\int_0^1\frac{x^{5/6}}{(1-x)^{1/6}\,(1+2\,x)^{4/3}}\log\left(\frac{1+2x}{x\,(1-x)}\right)\,dx$ Is it possible to find a closed form ...
31
votes
2answers
1k views

Closed form for improper definite integral involving trig functions and exponentials?

Is it possible to calculate the following definite integral in a closed form? $$ \int_0^\infty \left| \sin x \cdot \sin (\pi x) \right| e^{-x} \, dx$$
30
votes
2answers
623 views

A closed form of $\int_0^\infty\frac{\sqrt[\phi]{x}\ \arctan x}{\left(x^\phi+1\right)^2}dx$

Is it possible to evaluate the following integral in a closed form? $$\int_0^\infty\frac{\sqrt[\phi]{x}\ \arctan x}{\left(x^\phi+1\right)^2}dx,$$ where $\phi$ is the golden ratio: ...
30
votes
2answers
1k views

Integral $\int_0^1\frac{1-x^2+\left(1+x^2\right)\ln x}{\left(x+x^2\right)\ln^3x}dx$

I'm struggling with this integral $$I=\int_0^1\frac{1-x^2+\left(1+x^2\right)\ln x}{\left(x+x^2\right)\ln^3x}dx.\tag1$$ Mathematica could not evaluate it in a closed form. Its numeric value is ...
30
votes
1answer
992 views

Prove $\int_0^1\frac{x^2-2\,x+2\ln(1+x)}{x^3\,\sqrt{1-x^2}}\mathrm dx=\frac{\pi^2}8-\frac12$

How can I prove the following identity? $$\int_0^1\frac{x^2-2\,x+2\ln(1+x)}{x^3\,\sqrt{1-x^2}}\mathrm dx=\frac{\pi^2}8-\frac12$$
30
votes
2answers
1k views

Are there other cases similar to Herglotz's integral $\int_0^1\frac{\ln\left(1+t^{4+\sqrt{15}}\right)}{1+t}\ \mathrm dt$?

This post of Boris Bukh mentions amazing Gustav Herglotz's integral $$\int_0^1\frac{\ln\left(1+t^{\,4\,+\,\sqrt{\vphantom{\large A}\,15\,}\,}\right)}{1+t}\ \mathrm ...
30
votes
3answers
1k 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, ...
30
votes
1answer
647 views

$\int_0^1\arctan\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{x}{64}\right)\,\mathrm dx$

I need help with calculating this integral: $$\int_0^1\arctan\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{x}{64}\right)\,\mathrm dx,$$ where ...
30
votes
3answers
875 views

$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$

Please help me to solve this integral: $$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx.$$ I managed to calculate an indefinite integral of the left part: $$\int\frac{\cos x}{\sin ...