Closed form solution for $int_0^1 frac{ln^2(x)ln^2(1-x)}{x(1-x)} dx$ using Bose Integral The...
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Closed form solution for $int_0^1 frac{ln^2(x)ln^2(1-x)}{x(1-x)} dx$ using Bose Integral
The 2019 Stack Overflow Developer Survey Results Are InIntegral of $int_0^inftyfrac{1}{e^{x}-x} dx$Closed-form of integral $int_0^1 int_0^1 frac{arcsinleft(sqrt{1-s}sqrt{y}right)}{sqrt{1-y} cdot (sy-y+1)},ds,dy $Struggling to find a Closed Form for an IntegralClosed-form expression for integral involving exponential of cosines?Closed-form solution for integralDoes this integral have a closed form or asymptotic expansion? $int_0^infty frac{sin(beta u)}{1+u^alpha} du$Is there a closed form for $int_0^infty frac{1-cos(tx)}{e^t-1}dt$?Closed form solution for $int_0^{infty } frac{sin ({n}/{x})}{e^{2 pi x}-1} , dx$Is there a closed form solution of this definite integralClosed-form expression for an integral
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I'm looking for the solution to the following integral, but by using the Bose integral:
$$int_0^1 frac{ln^2(x)ln^2(1-x)}{x(1-x)} dx$$
I got to this form when looking for a solution to the following, $I=int_0^{pi/2} frac{ln^2(sin x)ln^2(cos x)}{sin x cos x}dx$ , after some reworking. I know this specific integral has a post about it already but to be honest I can't find it. As a result, I do not recall the closed for solution, and wolfram doesn't seem to know either. As I mentioned before, I'm looking for a solution that utilizes the Bose Integral, which is as follows $$Gamma(s)zeta(s)=int_0^infty frac{x^{s-1}}{e^x-1}dx$$ I used this integral to show the solution to $$J =int_0^{pi/2} frac{ln(sin x)ln(cos x)}{tan x}dx = frac{1}{16}int_0^infty frac{x^{2}}{e^x-1}dx = frac{zeta(3)}{8}$$ and am wondering if a similar argument can be made about the initial integral to reach a closed for solution like the one above due to the similarities between I and J.
calculus integration definite-integrals riemann-zeta
edited Mar 21 at 15:20
clathratus
5,1141439
asked Mar 20 at 3:38
Suchetan DonthaSuchetan Dontha
14812
$endgroup$
add a comment |
$begingroup$
I'm looking for the solution to the following integral, but by using the Bose integral:
$$int_0^1 frac{ln^2(x)ln^2(1-x)}{x(1-x)} dx$$
I got to this form when looking for a solution to the following, $I=int_0^{pi/2} frac{ln^2(sin x)ln^2(cos x)}{sin x cos x}dx$ , after some reworking. I know this specific integral has a post about it already but to be honest I can't find it. As a result, I do not recall the closed for solution, and wolfram doesn't seem to know either. As I mentioned before, I'm looking for a solution that utilizes the Bose Integral, which is as follows $$Gamma(s)zeta(s)=int_0^infty frac{x^{s-1}}{e^x-1}dx$$ I used this integral to show the solution to $$J =int_0^{pi/2} frac{ln(sin x)ln(cos x)}{tan x}dx = frac{1}{16}int_0^infty frac{x^{2}}{e^x-1}dx = frac{zeta(3)}{8}$$ and am wondering if a similar argument can be made about the initial integral to reach a closed for solution like the one above due to the similarities between I and J.
calculus integration definite-integrals riemann-zeta
edited Mar 21 at 15:20
clathratus
5,1141439
asked Mar 20 at 3:38
Suchetan DonthaSuchetan Dontha
14812
$endgroup$
$begingroup$
Differentiating beta function $B(x,y)$ twice with respect to x and twice with respect to y and then checking the value at x=y=0 might also help I guess..
$endgroup$
– Darkrai
Mar 20 at 7:59
1
$begingroup$
Heuristically, your integral is $16zeta(5)-8zeta(2)zeta(3)$
$endgroup$
– FDP
Mar 20 at 14:43
add a comment |
$begingroup$
I'm looking for the solution to the following integral, but by using the Bose integral:
$$int_0^1 frac{ln^2(x)ln^2(1-x)}{x(1-x)} dx$$
I got to this form when looking for a solution to the following, $I=int_0^{pi/2} frac{ln^2(sin x)ln^2(cos x)}{sin x cos x}dx$ , after some reworking. I know this specific integral has a post about it already but to be honest I can't find it. As a result, I do not recall the closed for solution, and wolfram doesn't seem to know either. As I mentioned before, I'm looking for a solution that utilizes the Bose Integral, which is as follows $$Gamma(s)zeta(s)=int_0^infty frac{x^{s-1}}{e^x-1}dx$$ I used this integral to show the solution to $$J =int_0^{pi/2} frac{ln(sin x)ln(cos x)}{tan x}dx = frac{1}{16}int_0^infty frac{x^{2}}{e^x-1}dx = frac{zeta(3)}{8}$$ and am wondering if a similar argument can be made about the initial integral to reach a closed for solution like the one above due to the similarities between I and J.
calculus integration definite-integrals riemann-zeta
edited Mar 21 at 15:20
clathratus
5,1141439
asked Mar 20 at 3:38
Suchetan DonthaSuchetan Dontha
14812
$endgroup$
I'm looking for the solution to the following integral, but by using the Bose integral:
$$int_0^1 frac{ln^2(x)ln^2(1-x)}{x(1-x)} dx$$
I got to this form when looking for a solution to the following, $I=int_0^{pi/2} frac{ln^2(sin x)ln^2(cos x)}{sin x cos x}dx$ , after some reworking. I know this specific integral has a post about it already but to be honest I can't find it. As a result, I do not recall the closed for solution, and wolfram doesn't seem to know either. As I mentioned before, I'm looking for a solution that utilizes the Bose Integral, which is as follows $$Gamma(s)zeta(s)=int_0^infty frac{x^{s-1}}{e^x-1}dx$$ I used this integral to show the solution to $$J =int_0^{pi/2} frac{ln(sin x)ln(cos x)}{tan x}dx = frac{1}{16}int_0^infty frac{x^{2}}{e^x-1}dx = frac{zeta(3)}{8}$$ and am wondering if a similar argument can be made about the initial integral to reach a closed for solution like the one above due to the similarities between I and J.
calculus integration definite-integrals riemann-zeta
calculus integration definite-integrals riemann-zeta
edited Mar 21 at 15:20
clathratus
5,1141439
asked Mar 20 at 3:38
Suchetan DonthaSuchetan Dontha
14812
edited Mar 21 at 15:20
clathratus
5,1141439
asked Mar 20 at 3:38
Suchetan DonthaSuchetan Dontha
14812
edited Mar 21 at 15:20
clathratus
5,1141439
edited Mar 21 at 15:20
clathratus
5,1141439
edited Mar 21 at 15:20
clathratus
5,1141439
5,1141439
asked Mar 20 at 3:38
Suchetan DonthaSuchetan Dontha
14812
asked Mar 20 at 3:38
Suchetan DonthaSuchetan Dontha
14812
asked Mar 20 at 3:38
Suchetan DonthaSuchetan Dontha
14812
14812
$begingroup$
Differentiating beta function $B(x,y)$ twice with respect to x and twice with respect to y and then checking the value at x=y=0 might also help I guess..
$endgroup$
– Darkrai
Mar 20 at 7:59
1
$begingroup$
Heuristically, your integral is $16zeta(5)-8zeta(2)zeta(3)$
$endgroup$
– FDP
Mar 20 at 14:43
add a comment |
$begingroup$
Differentiating beta function $B(x,y)$ twice with respect to x and twice with respect to y and then checking the value at x=y=0 might also help I guess..
$endgroup$
– Darkrai
Mar 20 at 7:59
1
$begingroup$
Heuristically, your integral is $16zeta(5)-8zeta(2)zeta(3)$
$endgroup$
– FDP
Mar 20 at 14:43
$begingroup$
Differentiating beta function $B(x,y)$ twice with respect to x and twice with respect to y and then checking the value at x=y=0 might also help I guess..
$endgroup$
– Darkrai
Mar 20 at 7:59
$begingroup$
Differentiating beta function $B(x,y)$ twice with respect to x and twice with respect to y and then checking the value at x=y=0 might also help I guess..
$endgroup$
– Darkrai
Mar 20 at 7:59
1
1
$begingroup$
Heuristically, your integral is $16zeta(5)-8zeta(2)zeta(3)$
$endgroup$
– FDP
Mar 20 at 14:43
$begingroup$
Heuristically, your integral is $16zeta(5)-8zeta(2)zeta(3)$
$endgroup$
– FDP
Mar 20 at 14:43
add a comment |
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$begingroup$
Differentiating beta function $B(x,y)$ twice with respect to x and twice with respect to y and then checking the value at x=y=0 might also help I guess..
$endgroup$
– Darkrai
Mar 20 at 7:59
1
$begingroup$
Heuristically, your integral is $16zeta(5)-8zeta(2)zeta(3)$
$endgroup$
– FDP
Mar 20 at 14:43