24 votes
Accepted

Prove $\int_0^1\frac{1}{\sqrt{1-x^2}}\arccos\left(\frac{3x^3-3x+4x^2\sqrt{2-x^2}}{5x^2-1}\right)\mathrm dx=\frac{3\pi^2}{8}-2\pi\arctan\frac12$.

The starting point strategy (optional) Beside this integral, there were already two more related integrals posted here before, namely: $\int_0^\frac{\pi}{3}\arccos(2\sin^2 x-\cos x)dx$ and $\int_{\...
Zacky's user avatar
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14 votes

How to determine the value of $\displaystyle f(x) = \sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n$?

In my opinion, getting the integral representation is too little for a post, but in order to put some meat on the bones, we can find a full asymptotics using this formula. $$S(x)=\sum_{n=1}^\infty\...
Svyatoslav's user avatar
  • 15.8k
14 votes
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Prove $\int_0^\pi\arcsin(\frac14\sqrt{8-2\sqrt{10-2\sqrt{17-8\cos x}}})dx=\frac{\pi^2}{5}$.

Applying $\arcsin x=\frac12\arccos(1-2x^2)$ followed by $\arccos x=\frac12\arccos(2x^2-1)$ yields: $$ \arcsin\left(\frac14\sqrt{8-2\sqrt{10-2\sqrt{17-8\cos x}}}\right)=\frac14\arccos\left(\frac14-\...
Zacky's user avatar
  • 27.8k
9 votes

Show that $\int_{0}^{1} \frac{\tan^{-1}(x^2)}{\sqrt{1 - x^2}} \, dx = \frac{1}{2}\pi \tan^{-1}\left(\sqrt{\frac{1}{\sqrt{2}} - \frac{1}{2}}\right)$

Let $I(a)= \int_0^{\pi/2} \tan^{-1}(\sinh a \sin^2t)dt$. Then \begin{align} I’(a)=&\int_0^{\pi/2}\frac{\cosh a \sin^2t}{1+\sinh^2a\sin^4t}dt=\frac\pi4\text{sech}\frac a2 \end{align} and \begin{...
Quanto's user avatar
  • 98.1k
9 votes
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Find the closed form of $_3F_2(\frac{1}{4},\frac{3}{4},\frac{5}{4};\frac{3}{2},\frac{7}{4};1)$

Let $\mathcal{H}$ denote the value of the following generalized hypergeometric function at unity: $$\mathcal{H}:={_3F_2}{\left(\frac14,\frac34,\frac54;\frac32,\frac74;1\right)}.$$ Using Euler's ...
David H's user avatar
  • 30k
7 votes
Accepted

How to represent $x^n$ as a sum of $P_k:= (x)(x-1)\dots(x-k+1)$?

This is related to Stirling numbers of the first kind and the second kind. https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind https://en.wikipedia.org/wiki/...
Vezen BU's user avatar
  • 2,016
6 votes

Closed form for this generalisation of the gamma function. $f(x+1)=f(x)g(x+1) $

There is a meaningful formula if, for some natural number $n$, $$ \lim_{x \to \infty} \frac{d^n}{dx^n}\ln(g(x)) = 0. $$ Under this condition, the solution is $$ f(x) = f(0) \lim_{N \to \infty}\left(\...
Polygon's user avatar
  • 1,889
5 votes

How to determine the value of $\displaystyle f(x) = \sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n$?

This is not a complete solution but it gives only the result: a complete formula for a sum generalizing that of the OP. The development was initiated by the results of @Svyatoslav and completed to ...
Dr. Wolfgang Hintze's user avatar
5 votes
Accepted

Closed form for $\int_0^{\pi/2}\arctan\left(\frac12\sin x\right)\mathrm dx$?

\begin{align} &\int_0^{\pi/2}\tan^{-1}\left(\frac12\sin x\right)dx\\ = &\int_0^{\pi/2}\int_0^{\frac12} \frac {\sin x}{(1+y^2 \sin^2x)}dy \ dx\\=&\int_0^{\frac12} \frac {\sinh^{-1}y}{y \...
Quanto's user avatar
  • 98.1k
4 votes

Closed form for this generalisation of the gamma function. $f(x+1)=f(x)g(x+1) $

One can plug in $x=0,1,\dots$ and find a pattern. Also, call $f(0)=c$: $$f(x+1)=g(x+1)f(x)\\x=0:f(1)=g(1)f(0)=c\,g(1)\\x=1:f(2)=g(2)f(1)=c\,g(1)g(2)\\x=2:f(3)=g(3)f(2)=c\,g(3)g(2)g(1)\\\vdots$$ One ...
Тyma Gaidash's user avatar
3 votes

Show that $\int_{0}^{1} \frac{\tan^{-1}(x^2)}{\sqrt{1 - x^2}} \, dx = \frac{1}{2}\pi \tan^{-1}\left(\sqrt{\frac{1}{\sqrt{2}} - \frac{1}{2}}\right)$

Trying to work the antiderivative. $$\tan ^{-1}\left(x^2\right)=\sum_{n=0}^\infty \frac{(-1)^n }{2 n+1}\,x^{4 n+2}$$ $$I_n=\int \frac{x^{4 n+2}}{\sqrt{1-x^2}}\,dx=\frac{x^{4 n+3} }{4 n+3}\,\,\, _2F_1\...
Claude Leibovici's user avatar
3 votes

Find the closed form of $_3F_2(\frac{1}{4},\frac{3}{4},\frac{5}{4};\frac{3}{2},\frac{7}{4};1)$

In the book on special functions by Askey, Roy and Gasper, you will find the Thomae formula about $_3F_2(a,b,c;d,e;1).$ I have checked and your $abcde$ leads through this formula to $K\times \ _3F_2(...
Letac Gérard's user avatar
3 votes

Simplify in closed-form $\sum_n P_n(0)^2 r^n P_n(\cos \theta)$

To express the series as an integral, one can use the generating function for the product of two Legendre polynomials derived by Leonard C. Maximon here: \begin{equation} \sum_{p=0}^\infty t^{p}P_p(\...
Paul Enta's user avatar
  • 14.1k
3 votes

compute the following integral in closed form : $\int_0^{\frac{π}{2}}\frac{x}{(1+\sqrt 2)\sin^{2}(x)+8\cos^{2} x}dx$

Consider $$I := \int_0^{\frac\pi2} \frac{x \,dx}{\alpha^2 \sin^2 x + \beta^2 \cos^2 x}.$$ Changing variables to $$u = \tan x$$ transforms the integral to $$\int_0^\infty \frac{\arctan u \,du}{\alpha^2 ...
Travis Willse's user avatar
3 votes

Compute $\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx$

I assume we're using the principal branch of the trilogarithm, and we're integrating on the lower side of the branch cut (which is what Mathematica assumes). Let's integrate the function $$f(z) = \...
Random Variable's user avatar
3 votes

Conjecture ${\large\int}_0^\infty\left[\frac1{x^4}-\frac1{2x^3}+\frac1{12\,x^2}-\frac1{\left(e^x-1\right)x^3}\right]dx=\frac{\zeta(3)}{8\pi^2}$

Utilize the known integral $$\frac1\pi\int_0^\infty \frac{\sin\frac{xt}{2\pi}}{e^t-1}dt = \frac12 \coth\frac x2-\frac1x$$ to evaluate \begin{align} &\int_0^\infty\left[\frac1{x^4}-\frac1{2x^3}+\...
Quanto's user avatar
  • 98.1k
2 votes

How to Evaluate $\sum_{n=0}^{\infty}\frac{(-1)^n(4n+1)(2n)!^3}{2^{6n}n!^6}$

We prove in this answer Bauer's series that was known since 1859. The original proof uses Fourier-Legendre expansions as I know. This series appears in one of the letters that Ramanujan sent to Hardy. ...
User's user avatar
  • 275
2 votes

compute the following integral in closed form : $\int_0^{\frac{π}{2}}\frac{x}{(1+\sqrt 2)\sin^{2}(x)+8\cos^{2} x}dx$

Let $a=\sqrt{8(\sqrt2-1)}$\begin{align} &\int_0^{\frac{π}{2}}\frac{x}{(1+\sqrt 2)\sin^{2} (x)+8\cos^{2} x}dx\\ \overset{ibp}=& \ \frac a8\int_0^{\frac{π}{2}}\cot^{-1}\frac{\tan x}{a}dx = \...
Quanto's user avatar
  • 98.1k
2 votes

Simplify in closed-form $\sum_n P_n(0)^2 r^n P_n(\cos \theta)$

Not a complete answer, but too long for a comment. For $S(\theta)$, we only need to sum over even indices since $P_n(0)=0$ for odd $n$. Additionally, as seen here, $$P_{2n}(0)=(-1)^n \frac{(2n-1)!!}{(...
Stefan Lafon's user avatar
  • 12.3k
2 votes
Accepted

How do I calculate the closed form of $\sum_{k=2}^\infty kx^{k-2}$

Since, if $x\in(-1,1)$,$$\frac1{1-x}=\sum_{k=0}^\infty x^k,$$you have that, on that interval$$\frac1{(1-x)^2}=\sum_{k=1}^\infty kx^{k-1};$$in other words,$$\frac1{(1-x)^2}-1=\sum_{k=2}^\infty kx^{k-1}....
José Carlos Santos's user avatar
2 votes

Power series for $\sum_{n=0}^\infty(-1)^n/n!^s$ (around $s=0$)

Denoting $\,S(s)=\displaystyle \sum_{n=0}^\infty\frac{(-1)^n}{n!^s}=\sum_{n=0}^\infty(-1)^ne^{-s\ln\Gamma(n+1)}$ we can use the Lindelöf summation formula for alternating series (also, for example, ...
Svyatoslav's user avatar
  • 15.8k
2 votes

Show that $\int_{0}^{1} \frac{\tan^{-1}(x^2)}{\sqrt{1 - x^2}} \, dx = \frac{1}{2}\pi \tan^{-1}\left(\sqrt{\frac{1}{\sqrt{2}} - \frac{1}{2}}\right)$

Let’s start from $$I=\int_0^{\frac{\pi}{2}} \tan ^{-1}\left(\sin ^2 t\right) d t$$ with its parametrised integral $$\int_0^{\frac{\pi}{2}} \tan ^{-1}\left(a\sin ^2 t\right) d t$$ Differentiating w.r.t ...
Lai's user avatar
  • 20.6k
2 votes

Closed form for $\int_0^{\pi/2}\arctan\left(\frac12\sin x\right)\mathrm dx$?

The antiderivative is very complicated but using the bounds $$I=\int_0^{\frac \pi 2}\arctan\left(\frac{\sin (x)}2\right)\mathrm dx=\frac{1}{12} \left(\pi ^2-6 \sinh ^{-1}(2)\,\, \text{csch}^{-1}(2)\...
Claude Leibovici's user avatar
1 vote

Closed form for $\int_0^{\pi/2}\arctan\left(\frac12\sin x\right)\mathrm dx$?

Partial answer $$ I(a)=\int_0^{\pi/2}\tan^{-1}\left(a\sin(x)\right)\;dx\implies I'(a)=\int_0^{\pi/2}\frac{\sin(x)}{{1+a^2\sin^2x}}dx=\int_0^{\pi /2}\frac{\sin x}{1+a^2-a^2\cos^2(x)}dx$$ Letting $t=\...
Masd's user avatar
  • 608
1 vote
Accepted

Integral in terms of Hypergeometric function

Using only Mathematica: $$\int _0^1\int _0^1\left(1+c^2 v^2\right)^{-s} u^{1-2 s} (1-u v)^{s-1}dvdu=\mathcal{M}_q^{-1}\left[\int _0^1\int _0^1\mathcal{M}_c\left[\left(1+c^2 v^2\right)^{-s} u^{1-2 s}...
Mariusz Iwaniuk's user avatar
1 vote

Integral in terms of Hypergeometric function

This is not an answer From a computation point of view, why not to stay with $$I=\int_0^1 \frac { B_v(2(1- s),s)} {v^{2(1- s)} \left(1+c^2 v^2\right)^s }\,dv$$ which does not seem to make too much ...
Claude Leibovici's user avatar
1 vote

How to represent $x^n$ as a sum of $P_k:= (x)(x-1)\dots(x-k+1)$?

Maybe this can be another way to solve the problem, let me know if you agree with me. You can observe that $$P_k(j)=0$$ for each $j\leq k-1$, and that $$P_k(j)=j(j-1)\cdots (j-k+1)=\binom{j}{k}k!$$ ...
Federico Fallucca's user avatar
1 vote

Compute in closed form $\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$

I would like to consider a generalization of this problem. Define the function $\mathcal{I}:\mathbb{R}\times[-1,1]\rightarrow\mathbb{R}$ via the definite integral $$\mathcal{I}{\left(a,b\right)}:=\...
David H's user avatar
  • 30k

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