New answers tagged closed-form
1
vote
Nice Integral : $\int^1_0 \arctan\left({\frac{1-x}{1+x}}\right)\arctan(x)\frac{1}{1-x}dx$
$$\begin{align*}
I &= \int_0^1 \arctan \frac{1-x}{1+x} \cdot \arctan x \cdot \frac{dx}{1-x} \\
&= \int_0^1 \frac{\left(\frac\pi4-\arctan x\right) \arctan x}{1-x} \, dx \\
I_1 &= \int_0^\...
- 21.2k
5
votes
Nice Integral : $\int^1_0 \arctan\left({\frac{1-x}{1+x}}\right)\arctan(x)\frac{1}{1-x}dx$
\begin{align}
I=&\int^1_0 \frac{\tan^{-1}{\frac{1-x}{1+x}}\tan^{-1}x}{1-x}dx\\
\overset{ibp}=&
\int^1_0 \frac{\ln(1-x)\left( \tan^{-1}{\frac{1-x}{1+x}}-\tan^{-1}x\right)}{1+x^2}\overset{x\to ...
- 102k
3
votes
Accepted
Nice Integral : $\int^1_0 \arctan\left({\frac{1-x}{1+x}}\right)\arctan(x)\frac{1}{1-x}dx$
Note that $\tan^{-1}(\frac{1-x}{1+x})=\frac{\pi}{4}-\tan^{-1}(x)$. With partial fraction $\frac{1}{x(1+x)}=\frac{1}{x}-\frac{1}{1+x}$
We have
$$I=\int_0^1\tan^{-1}\left(\frac{1-x}{1+x}\right)\tan^{-1}(...
- 754
0
votes
Integrate $\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$
Since @Vladimir_Reshetnikov gave an approach by transforming the integral into $$\int_0^{\infty}\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}\right)\, dy$$
I kinda ...
- 754
0
votes
$I(x) = -\int_0^1 \frac{1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)\,dz$
Let $x=\cos2a$ for notational convenience, i.e.
$$I(a) = \int_0^1 \frac{1}{z}\ln\bigg(\frac{1-z\cos2a+ \sqrt{1-2z \cos2a+ z^2}}{2}\bigg) dz$$
and substitute $z-\cos2a =\sin2a\sinh t$ to remove the ...
- 102k
1
vote
$\int_0^1 \sqrt{\frac{2-x^4}{1-x^4}}\, dx$ in terms of the gamma function or elliptic integrals
Too long for comment and not really an answer to the central question, but here is how we can recover the hypergeometric result:
$$\begin{align*}
I &= \int_0^1 \sqrt{\frac{2-x^4}{1-x^4}} \, dx \\
&...
- 21.2k
3
votes
Accepted
how to evaluate $\int_{0}^{1} \int_{0}^{1} \frac{\ln(1 - xy) \cdot \text{Li}_{4}(1 - x)}{x(1 - x)(1 - xy)} \,dy\,dx$
Complete solution $100\%$ Euler sum free!
$a>0,b\geq 0$ integers,
\begin{align*}R_a&=\int_0^1\frac{\ln^ax}{1-x}dx,L_{a,b}=\int_0^1\frac{\ln^ax}{1-x}\left(\int_0^x\frac{\ln^bt}{1-t}dt\right)dx\\
...
- 14.2k
6
votes
How to evaluate $\int_0^1 \frac{\arctan(x)}{x} \ln^2(1 - x) \, dx$
We need the classical result,
$$\int_{0}^{1}x^{n-1}\log^2(1-x)\textrm{d}x=\frac{2}{n}\sum\limits_{k=1}^{n}\frac{H_{k}}{k}=\frac{H_{n}^2+H_n^{(2)}}{n}, \tag1$$
also given in (Almost) Impossible ...
- 5,862
1
vote
How to evaluate $\int_0^1 \frac{x}{1+x^2} \ln ^2\left(\frac{x}{1-x}\right) d x$
Answer using complex analysis
$$I=\int_0^1\frac{x}{x^2+1}\ln^2\left(1-\frac{1}{x}\right)dx=\int_0^\infty\frac{\ln^2x}{(x+1)(x^2+2x+2)}dx$$
Define $f(z)=\frac{\ln^3z}{(z+1)(z^2+2z+2)}$ and integrate $f(...
- 2,542
2
votes
How to evaluate $\int_0^1 \frac{\arctan(x)}{x} \ln^2(1 - x) \, dx$
If you use
$$\log ^2(1-x)=\sum_{n=0}^\infty \frac {a_n}{b_n}\,x^{n+2}$$ where the $a_n$ and $b_n$ correspond respectively to sequences $A002547$ and $A002548$ in $OEIS$ (all of them are positive).
If
$...
- 270k
4
votes
How to evaluate $\int_0^1 \frac{\arctan(x)}{x} \ln^2(1 - x) \, dx$
Incomplete solution
Consider the parametrized integral with the expansion of $\frac{arctan x}{x}$, we have
$$
\begin{aligned}
I(a) & =\int_0^1 \frac{(1-x)^a \arctan x}{x} d x \\
& =\sum_{n=0}^{...
- 23.7k
1
vote
How to evaluate $\int_0^1 \frac{\arctan(x)}{x} \ln^2(1 - x) \, dx$
Here is an option, however, not a closed-form solution. Expand $\frac{\arctan(x)}{x}$ into a series, i.e.,
$$\frac{\arctan(x)}{x}=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{2n+1}.$$
Thus, you need to ...
- 1,569
4
votes
Find: $ \int_0^{\pi/2} \int_0^{\pi/2} \frac{\cos^2(x) \cos(t)}{(b^2 - a^2 \sin^2(x))(1 + \sin^2(t) \cos^2(x))} \ dt \ dx $
Let
$$J(t)= \int_0^{\pi/2} \frac{\cos x\tan^{-1}(\tan t\cos x)}{1-\sin^2s\sin^2x}dx
$$
along with
\begin{align}
J’(t)=& \int_0^{\pi/2} \frac{\sec^2s\sec^2t\cos^2x}{(1+\tan^2s\cos^2x) (1+\tan^2t\...
- 102k
3
votes
Find: $ \int_0^{\pi/2} \int_0^{\pi/2} \frac{\cos^2(x) \cos(t)}{(b^2 - a^2 \sin^2(x))(1 + \sin^2(t) \cos^2(x))} \ dt \ dx $
Assuming $0<a<b$, introduce and differentiate w.r.t. a parameter $k\in\Bbb R$:
$$\begin{align*}
J(k) &= \int_0^\tfrac\pi2 \frac{\cos x}{b^2-a^2\sin^2x} \arctan(k \cos x) \, dx \\[2ex]
J'(k) &...
- 21.2k
1
vote
How to evaluate $\int_0^1 \ln ^3(1+x) \ln (1-x) d x$?
$$I=\int_0^1 \ln ^4 t \frac{d t}{(1+t)^2}=\int_0^1 \frac{\ln ^4 t}{(1+t)^2} d t=\int_0^1\sum_{n=0}^\infty(-1)^n(n+1)t^n\ln^4tdt=\sum_{n=0}^\infty(-1)^n\frac{24}{(n+1)^4}=\frac{45}{2}\zeta(5).
$$
- 45.9k
3
votes
How to evaluate $\int_0^1 \frac{x}{1+x^2} \ln ^2\left(\frac{x}{1-x}\right) d x$
For the first problem
$$I=\int \frac{x}{1+x^2} \log ^2\left(\frac{x}{1-x}\right)\,dx$$
$$\frac{x}{1-x}=t \implies x=\frac{t}{t+1}\implies dx=\frac{dt}{(t+1)^2}$$
$$I=\int\frac{t }{2 t^3+4 t^2+3 t+1}\...
- 270k
2
votes
Accepted
How to evaluate $\int_0^\infty \tanh\left(\frac{\pi x}{2}\right) \left(\frac{1}{x} - \frac{x}{x^2 + y^2}\right) \, dx $
Let's consider
$$I(a,b)=\int_0^\infty \tanh\left(\frac{\pi x}{2}\right) \left(\frac{x}{x^2 + a^2}- \frac{x}{x^2 + b^2}\right) \, dx$$
then the initial integral $\,\displaystyle I_0=I(a=0, b=y)$. ...
- 16.7k
2
votes
How to evaluate $\int_0^\infty \tanh\left(\frac{\pi x}{2}\right) \left(\frac{1}{x} - \frac{x}{x^2 + y^2}\right) \, dx $
Concerning the last summation, since
$$s + (k+1)\pi= \pi \left(k+\frac{s+\pi }{\pi }\right)$$
$$\sum_{k=0}^\infty \frac{(-1)^k}{s + (k+1)\pi}=\frac 1{2\pi}\left(\psi \left(\frac{s+2 \pi }{2 \pi }\...
- 270k
1
vote
how to evaluate $\int_{0}^{1} \int_{0}^{1} \frac{\ln(1 - xy) \cdot \text{Li}_{4}(1 - x)}{x(1 - x)(1 - xy)} \,dy\,dx$
We want to find
$$I:=\int_0^1\frac{\text{Li}_4(x)\ln^2x}{x(1-x)^2}dx$$
We have $\frac{d}{dx}(\text{Li}_4(x))=\frac{\text{Li}_3(x)}{x}$, and according to WolframAlpha, we have
$$\int \frac{\ln^2x}{x(1-...
- 146k
3
votes
How to evaluate $\int_{0}^{\frac{\pi}{2}} \left[\ln\left(\frac{e^{-x^2}}{\cos(x)} \left(1 + \cos(4x)\right)\right)\right]^2 \, dx $
First, simplify $\dfrac{1+\cos(4x)}{\cos x}=2\cos^2(2x)$, then expand the integrand to
$$\log^22 - 2(\log2) x^2 + x^4 + 2\log 2\left[\log\left(\cos^2(2x)\right) - \log(\cos x)\right] \\
+ 2x^2 \left[\...
- 21.2k
4
votes
Remarkable logarithmic integral $\int_0^1 \frac{x \log ^2(x) \log (1-x)}{1+x^2} dx$
Not an answer, but a comment on the integral of type
$$I(a,b,c) = \int_0^1 \frac{x}{1+x^2} (\log x)^a (\log(1-x))^b (\log(1+x))^c dx $$
Visually, this is a level 4 integral, so in generally one only ...
8
votes
How to evaluate $\int_0^1 \frac{x}{1+x^2} \ln ^2\left(\frac{x}{1-x}\right) d x$
For a complex number $z\ne -1$, the following identity holds:
$$\int_0^1\frac{z\ln^{q}(\frac{x}{1-x})}{1+zx}dx=2(-1)^{q}q!\sum_{n=0}^{\lfloor\frac{q}{2}\rfloor}\frac{\ln^{q-2n+1}(1+z)}{(q-2n+1)!}\eta(...
- 26.3k
2
votes
Evaluate $\int^{\infty}_{0} \arctan\left({\frac{a^2}{x^2}}\right)\cos(bx)dx$
WLOG assume $a,b>0$. Sketch of a solution.
Integration by parts with $dv= \cos(bx)$. The boundary terms vanish.
The new integral is
$$\frac{2a^2}{b}\int_{0}^{\infty}\frac{x\sin(bx)}{a^4+x^4}\,dx$$...
- 8,047
4
votes
Accepted
Evaluate $\int^{\infty}_{0} \arctan\left({\frac{a^2}{x^2}}\right)\cos(bx)dx$
As already observed by J.G., you special integral has the right result $\pi e^{-\vert b\vert}\frac{\sin(\vert b\vert)}{\vert b\vert}$.
I suggest to use a change of variable to use the integral that ...
- 9,150
1
vote
Closed form of the following sum of binomial coefficients.
Let us define the sequence $a(n)$ by
$$
a(n) = \sum_{\begin{array}{cc}
1\le k\le n-1\\
gcd(k,n)=1
\end{array}}{n\choose k}
$$
It is perhaps worth mentioning that the sequence $a(n)$ is
...
- 133k
3
votes
Closed form of the following sum of binomial coefficients.
This is sequence A056188 in OEIS (Encyclopedia of Integer Sequences). No closed-form formula is mentionned in the corresponding article.
Here is a (log-scale) representation of the sequence (blue ...
- 84.5k
0
votes
Show that $\int_{0}^{\frac{\pi}{2}} \log(1 + \tan^4 x) \cos^2 x \, dx = \frac{\pi \log(6 + 4\sqrt{2})}{4} - \frac{\pi}{2}$
\begin{aligned}
\int_0^{\pi/2} \ln(1+\tan^4 x) \cos^2 x \, dx &= \left(\int_0^{\pi/2} \frac{\ln(1+\tan^4 x)}{(1+\tan^2 x)^2} \, d\tan x \Big|_{0}^{\pi/2}\right) \Bigg|_{x \to \tan x} \\
&= \...
- 5,196
1
vote
How to evaluate $\int_0^1 \frac{x}{1+x^2} \ln ^2\left(\frac{x}{1-x}\right) d x$
Incomplete solution.
Let $n\geq 2$, an integer.
\begin{align}J_n&=\int_0^1 \frac{x\ln^n\left(\frac{x}{1-x}\right)}{1+x^2}dx\\
&\overset{u=\frac{1-x}{1+x}}=\int_0^1 \frac{\ln^n\left(\frac{1-y}{...
- 14.2k
1
vote
Accepted
Closed form : $\int^{\infty}_{1}\frac{1}{x^4-2x^2+2} dx $
Too long for a comment.
As you wrote $$I=\int^{\infty}_{1}\frac{1}{x^4-2x^2+2}\, dx=\frac i 2 \int^{\infty}_{1}\Bigg(\frac{1}{x^2-(1-i)}-\frac{1}{x^2-(1+i)} \Bigg) \,dx$$ Then
$$I=\frac{i}{2} \left(...
- 270k
3
votes
How to evaluate $ \int_0^\infty \ln^3(1 - e^{-\pi x}) \tanh(\pi x) \, dx $
\begin{align}
&\int_0^\infty \ln^3(1 - e^{-\pi x}) \tanh(\pi x) dx\\
=& \ \frac{1}{\pi} \int_0^1 \frac{\ln^3 x\left(1 - (1 - x)^2\right)}{(1 + (1 - x)^2)(1-x)} \, dx \\
= &\ \frac{1}{\pi}...
- 102k
6
votes
How to evaluate $\int_0^1 \frac{x}{1+x^2} \ln ^2\left(\frac{x}{1-x}\right) d x$
You can check that your integral can be expressed as
$$\left.\dfrac{\mathrm d^2}{\mathrm da^2}\right|_{a=1}\int_0^1\dfrac{x^a}{(1+x^2)(1-x)^{a-1}}\,\mathrm d x$$
and according to WA
$$\int_0^1\dfrac{x^...
- 3,624
3
votes
Accepted
How to evaluate $ \int_0^\infty \ln^3(1 - e^{-\pi x}) \tanh(\pi x) \, dx $
Too Long for comment:
$$ \int_{0}^{1} \frac{ \ln^{n}(x)}{1-x} \ dx = (-1)^{n} n! \ \zeta(n+1)$$
$$I= \frac{1}{\pi} \int_0^1 \ln^3(x) \left(\frac{1}{1 - x} - \frac{2(1 - x)}{1 + (1 - x)^2}\right) \, ...
- 3,578
0
votes
Can we further simplify the closed solution to the Goat Problem?
Assuming that you isolated a simple root $z_0$ of the equation $f(z)=0$ in a domain $\mathcal D$ (such as a disk that contains it), that root can be obtained as
$$\dfrac{\displaystyle\oint_{\mathcal D}...
- 3,510
0
votes
Can we further simplify the closed solution to the Goat Problem?
Because there isimplied some complex trigonometric function algebra, use Mathematica
...
- 3,666
1
vote
How to evaluate $\int_{0}^{\frac{\pi}{2}} \frac{\cos(x)}{(1 + \sqrt{\sin(2x)})^n} \,dx$
\begin{align}
&\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(1 + \sqrt{\sin2x})^n} \,dx \\
= &\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos x + \sin x}{(1 + \sqrt{\sin2x})^n} \,dx
=\int_{\frac\...
- 102k
1
vote
How to evaluate $\int_{0}^{\frac{\pi}{2}} \left[\ln\left(\frac{e^{-x^2}}{\cos(x)} \left(1 + \cos(4x)\right)\right)\right]^2 \, dx $
Using ONLY Mathematica:
$$ I = \int_{0}^{\frac{\pi}{2}} \left[\ln\left(\frac{e^{-x^2}}{\cos(x)} \left(1 + \cos(4x)\right)\right)\right]^2 \, dx=-\frac{11 \pi \zeta (3)}{16}+\frac{\pi ^3}{4}+\frac{\...
- 4,562
3
votes
Accepted
How to evaluate $\int_{0}^{\frac{\pi}{4}} \frac{\ln(\sin x) \ln(\cos x)}{\tan(2x)} \sin(2x) \, dx $
$$
\begin{aligned}
& \quad \int_0^{\frac{\pi}{4}} \frac{\ln (\sin x) \ln (\cos x)}{\tan (2 x)} \sin (2 x) d x \\&=\int_0^{\frac{\pi}{4}} \cos (2 x) \ln (\sin x) \ln (\cos x) d x \\
& =\...
- 23.7k
1
vote
How can I show that $\int_0^{\frac{\pi}{2}}\sin\left(\frac{x}{2}\right) \text{arctanh}\left(\sin(2x)\right) dx$ has this value?
More generally, the integral $$\int_{0}^{\pi/2} \sin(ax) \operatorname{artanh}(\sin 2x) \, \mathrm dx $$ can be expressed in terms of the digamma function.
Assume that $0 < a< 2$.
For $|r| &...
- 43k
6
votes
How to evaluate $\int_{0}^{\frac{\pi}{4}} \frac{\ln(\sin x) \ln(\cos x)}{\tan(2x)} \sin(2x) \, dx $
Alternatively
\begin{align}
&\int_{0}^{\frac{\pi}{4}} \frac{\ln(\sin x) \ln(\cos x)}{\tan2x} \sin2x\, dx\\
= &\ \frac12\int_{0}^{\frac{\pi}{4}} {\ln(\sin x) \ln(\cos x)}d\left(\sin 2x\right)\\
...
- 102k
4
votes
Closed form for$ \int_0^1 x^n \sqrt{1-x^2}\ln x\ dx $
The integral $I_n= \int_0^1 x^n \sqrt{1-x^2}\ln x\ dx $ has the closed form
$$I_n=\frac{(n-1)!!}{(n+2)!!}\bigg(\ln2 + \frac{(-1)^n}{n+2}+\sum_{k=1}^{n} \frac{(-1)^{k}}{k}\bigg) \bigg( \sin^2\frac{n \...
- 102k
2
votes
How to find closed form $\int^1_0\frac{\arctan(x)}{1+3x^2}dx$
More generally
\begin{align}\int_0^{1}& \frac{\tan^{-1} x}{1+a^{2}x^2}{dx}
=\int_0^1 \int_0^1\frac x{(a^{2}x^2+1)(y^2x^2+1)}dy\ dx\\
&\>\>\>\>\> = \frac12\int_0^1 \frac{\ln \...
- 102k
1
vote
Accepted
How to find closed form $\int^1_0\frac{\arctan(x)}{1+3x^2}dx$
Hint
$$I=\int\frac{\log(1+y^2)}{3-y}\,dy=\int\frac{\log(y+i)}{3-y}\,dy+\int\frac{\log(y-i)}{3-y}\,dy$$
Use
$$\int \frac{\log(y+a)}{b-y}\,dy=-\log (a+y)\, \log \left(\frac{b-y}{a+b}\right)-\text{Li}_2\...
- 270k
3
votes
How to evaluate $\int_{0}^{\frac{\pi}{2}} \left[\ln\left(\frac{e^{-x^2}}{\cos(x)} \left(1 + \cos(4x)\right)\right)\right]^2 \, dx $
(NOTE: Too long for comment)
(NOTE: I have utilized "Possible Closed forms" from WA in $J_1,J_3$)
$$ I = \int_{0}^{\frac{\pi}{2}} \left[\ln\left(\frac{e^{-x^2}}{\cos(x)} \left(1 + \cos(4x)\...
- 3,578
3
votes
How to evaluate $\int_{0}^{1} \int_{0}^{1} \frac{4y \, da \, dy}{(y^2 + ay + 1)(y^2 - ay + 1)}$?
Ramanujan found that
$$\int_{0}^{\pi/2} \ln \left( \frac{2 + \sin x}{2 - \sin x} \right) \, dx \, = \, \color{blue}{\displaystyle\sum_{n=0}^\infty \displaystyle\frac{1}{\displaystyle\binom{2n}{n}\left(...
- 18.1k
3
votes
How to evaluate $\int_{0}^{\frac{\pi}{2}} \left[\ln\left(\frac{e^{-x^2}}{\cos(x)} \left(1 + \cos(4x)\right)\right)\right]^2 \, dx $
Partial answer
$$J_2 = \int\log\left(e^{-x^2}\right) \log\left(\frac{1 + \cos(4x)}{\cos(x)}\right) \, dx=
- \int x^2 \log\left(\frac{1 + \cos(4x)}{\cos(x)}\right) \, dx$$ can be computed using a ...
- 270k
0
votes
A closed form of $\int_0^\infty\frac{\sqrt[\phi]{x}\ \arctan x}{\left(x^\phi+1\right)^2}dx$
First, substituting $x \to x^\phi$:
$$\int_{0}^{\infty} \frac{x \arctan(x^\phi)}{(x^{\phi^2} + 1)^2} \, dx$$ $$=\phi
\int_{0}^{\infty} \frac{x^\phi \cdot \arctan(x^\phi)}{(x^{\phi + 1} + 1)^2} \, dx$$...
- 5,196
1
vote
Prove that $\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx =\frac{\pi e}{24} $
Similar to user153012’s solution
\begin{align*}
\int_{0}^{1} x^n \sin(\pi x) \, x^x \, (1 - x)^{1 - x} \, dx &= b_{n+2} \pi e, \\
b_1 &= \frac{1}{2}, \\
b_n &= \frac{1}{n} \left( \frac{1}{...
- 5,196
4
votes
Closed form for$ \int_0^1 x^n \sqrt{1-x^2}\ln x\ dx $
Noticing that
$$
I_n=\int_0^1 x^n \sqrt{1-x^2} \ln x d x=I^{\prime}(n)
$$
where
$$
I(a)=\int_0^1 x^a \sqrt{1-x^2} d x
$$
Via the substitution $x=\sin \theta$, we have
$$
\begin{aligned}
I(a) & =\...
- 23.7k
1
vote
Closed form for$ \int_0^1 x^n \sqrt{1-x^2}\ln x\ dx $
Making the substitution $x=\cos\theta$ you get
$$\int_0^{\frac{\pi}{2}} \cos^n(\theta)\sin^2\theta\ln(\cos\theta)\,\mathrm d\theta$$
Furthermore, we can substitute the Fourier series of $\ln(\cos\...
- 3,624
4
votes
Closed form for$ \int_0^1 x^n \sqrt{1-x^2}\ln x\ dx $
This is too long for a comment. Gradshteyn and Rhyzik's table of integrals provides the following expansions:
$$\int_0^1 x^{2n}\sqrt{1-x^2}\ln x\,dx=\frac{(2n-1)!!}{(2n+2)!!}\cdot\frac{\pi}{2}\left(\...
- 17.6k
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