product= $expleft[frac47mathrm G30pi+frac34right]left(frac11^113^313^13right)^1/20sqrtfrac37^7/6pisqrtfrac2pi$Integral: $int_0^pi/12 ln(tan x),dx$Proving the closed form $int_pi/20^3pi/20 ln tan x,,dx= - frac2G5$Integral $int_0^inftyexpleft(-sqrt2,x^2right),operatornameerfi(x),log(x),x^3,dx$Connection between integral expression and the factorial of infinityProve $intlimits_0^infty mathrmexp(-ax^2-fracbx^2) mathrmd x = frac12sqrtfracpia}mathrme^{-2sqrtab$A faster way to evaluate $int_1^inftyfracsqrt4+t^2t^3,mathrm dt$?Evaluate $int_0^infty fracx expleft-beta^2 x^2rightsinh left(fracpi x2right) mathrmd x$Show that $intlimits_0^fracpi24cos^2(x)log^2(cos x)~mathrm dx=-pilog 2+pilog^2 2-fracpi2+fracpi^312$Integrating $fracxsin x$ using infinite products and fraction decompositionProve $int_0^1fraclog(t^2-t+1)t^2-tmathrm dt=fracpi^29$Relationship between Catalan's constant and $pi$Alternative proof for $zetaleft(2,frac14right)=psi^(1)left(frac14right)=pi^2+8G$
I keep switching characters, how do I stop?
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product= $expleft[frac47mathrm G30pi+frac34right]left(frac11^113^313^13right)^1/20sqrtfrac37^7/6pisqrtfrac2pi$
Integral: $int_0^pi/12 ln(tan x),dx$Proving the closed form $int_pi/20^3pi/20 ln tan x,,dx= - frac2G5$Integral $int_0^inftyexpleft(-sqrt2,x^2right),operatornameerfi(x),log(x),x^3,dx$Connection between integral expression and the factorial of infinityProve $intlimits_0^infty mathrmexp(-ax^2-fracbx^2) mathrmd x = frac12sqrtfracpiamathrme^-2sqrtab$A faster way to evaluate $int_1^inftyfracsqrt4+t^2t^3,mathrm dt$?Evaluate $int_0^infty fracx expleft-beta^2 x^2rightsinh left(fracpi x2right) mathrmd x$Show that $intlimits_0^fracpi24cos^2(x)log^2(cos x)~mathrm dx=-pilog 2+pilog^2 2-fracpi2+fracpi^312$Integrating $fracxsin x$ using infinite products and fraction decompositionProve $int_0^1fraclog(t^2-t+1)t^2-tmathrm dt=fracpi^29$Relationship between Catalan's constant and $pi$Alternative proof for $zetaleft(2,frac14right)=psi^(1)left(frac14right)=pi^2+8G$
$begingroup$
$mathrm G$ is Catalan's constant.
I recently found the product
$$
alpha=prod_n=1^inftyfracE_n(frac12)E_n(frac712)E_n(frac120)E_n(frac1320)E_n(frac14)E_n(frac112)E_n(frac320)E_n(frac1120)=\
expleft[frac47mathrm G30pi+frac34right]sqrtfrac3391pisqrtfrac2pifracsqrt[5]11sqrt[3]7sqrt[5]frac3^313^3$$
(an alternate form of the product in the title)
Where $$E_n(x)=fracj(n+x)(en)^2xj(n-x)qquad xin(0,1)$$
and $j(x)=x^x$.
Could I have some numerical evidence, or better yet an alternate proof? My tools are limited to desmos, which cannot really handle infinite products. Thanks.
My Proof.
We define $$mathrm L(x)=frac1piint_0^pi xlog(sin t)dt$$
And we use $$sin t=tprod_ngeq1left(1-fract^2pi^2 n^2right)$$
To see that $$log(sin t)=log(t)+sum_ngeq1logfracpi^2n^2-t^2pi^2n^2$$
Then integrate both sides over $[0,x]$ to get
$$pimathrm L(x/pi)=x(log x-1)+sum_ngeq1xlogbigg(1-fracx^2pi^2n^2bigg)-2x+pi nlogfracpi n+xpi n-x$$
$$pimathrm L(x/pi)=logleft[fracj(x)e^xright]+sum_ngeq1logleft[fracj(pi n+x)(epi n)^2xj(pi n-x)right]$$
$xmapsto pi x$:
$$pimathrm L(x)=logleft[fracj(pi x)e^pi xright]+sum_ngeq1logleft[fracj(pi n+pi x)(epi n)^2pi xj(pi n-pi x)right]$$
$$mathrm L(x)=logleft[left(fracpieright)^xj(x)right]+sum_ngeq1log E_n(x)$$
Then we define $$U(x)=prod_ngeq1E_n(x)$$
To see that $$U(x)=left(fracepi xright)^xexpmathrm L(x)$$
Where we used $$sum_nlog(a_n)=logleft[prod_na_nright]$$
and the neat rules $$log(a^b)=log(e^blog a)=blog a$$
$$log(a)pm b=logleft(e^pm baright)$$
to simplify the expressions. Next, we define
$$P_mu,nu(a_1,a_2,dots,a_mu;b_1,b_2,dots,b_nu)=fracprod_i=1^mu U(a_i)prod_i=1^nu U(b_i)$$
And we see that
$$P_mu,nu(a_1,dots,a_mu;b_1,dots,b_nu)=prod_ngeq1fracprod_i=1^mu E_n(a_i)prod_i=1^nu E_n(b_i)$$
This gives $$P_1,1(x_1;x_2)=left(fracepiright)^x_1-x_2fracj(x_2)j(x_1)expleft[mathrm L(x_1)-mathrm L(x_2)right]$$
Then we define $$mathrmT(x)=frac1piint_0^pi xlog(tan t)dt=mathrm L(x)-mathrm L(x+1/2)-frac12log2$$
To get that
$$P_1,1left(x;x+frac12right)=sqrtfrac2pie,fracj(x+1/2)j(x)expmathrm T(x)$$
So we have
$$P_2,2left(x_1,x_2+frac12 ;x_2,x_1+frac12right)=fracj(x_1+1/2)j(x_2)j(x_2+1/2)j(x_1)expleft[mathrm T(x_1)-mathrm T(x_2)right]$$
Then using the identities
$$mathrm L(1/2)=-frac12log2$$
$$mathrm L(1/4)=fracmathrm G2pi-frac14log2$$
We get $$P_1,1left(frac12;frac14right)=frac1(2pi)^1/4expleft[fracmathrm G2pi+frac14right]tag1$$
From here, the identity
$$-mathrm T(1/12)=frac2mathrm G3pi$$
which gives
$$P_1,1left(frac712;frac112right)=sqrtfrac67pisqrt[6]7expleft[frac2mathrm G3pi+frac12right]tag2$$
Then from here, the identity
$$mathrm T(1/20)-mathrm T(3/20)=frac2mathrm G5pi$$
gives $$P_2,2left(frac120,frac1320;frac320,frac1120right)=left(fracj(11)j(3)j(13)right)^1/20expfrac2mathrm G5pitag3$$
Then multiplying $(1),(2),$ and $(3)$, we have the desired result, namely
$$P_4,4left(frac12,frac712,frac120,frac1320;frac14,frac112,frac320,frac1120right)=alpha$$
calculus integration alternative-proof infinite-product constants
asked Mar 13 at 17:43
clathratusclathratus
5,0601338
$endgroup$
add a comment |
$begingroup$
$mathrm G$ is Catalan's constant.
I recently found the product
$$
alpha=prod_n=1^inftyfracE_n(frac12)E_n(frac712)E_n(frac120)E_n(frac1320)E_n(frac14)E_n(frac112)E_n(frac320)E_n(frac1120)=\
expleft[frac47mathrm G30pi+frac34right]sqrtfrac3391pisqrtfrac2pifracsqrt[5]11sqrt[3]7sqrt[5]frac3^313^3$$
(an alternate form of the product in the title)
Where $$E_n(x)=fracj(n+x)(en)^2xj(n-x)qquad xin(0,1)$$
and $j(x)=x^x$.
Could I have some numerical evidence, or better yet an alternate proof? My tools are limited to desmos, which cannot really handle infinite products. Thanks.
My Proof.
We define $$mathrm L(x)=frac1piint_0^pi xlog(sin t)dt$$
And we use $$sin t=tprod_ngeq1left(1-fract^2pi^2 n^2right)$$
To see that $$log(sin t)=log(t)+sum_ngeq1logfracpi^2n^2-t^2pi^2n^2$$
Then integrate both sides over $[0,x]$ to get
$$pimathrm L(x/pi)=x(log x-1)+sum_ngeq1xlogbigg(1-fracx^2pi^2n^2bigg)-2x+pi nlogfracpi n+xpi n-x$$
$$pimathrm L(x/pi)=logleft[fracj(x)e^xright]+sum_ngeq1logleft[fracj(pi n+x)(epi n)^2xj(pi n-x)right]$$
$xmapsto pi x$:
$$pimathrm L(x)=logleft[fracj(pi x)e^pi xright]+sum_ngeq1logleft[fracj(pi n+pi x)(epi n)^2pi xj(pi n-pi x)right]$$
$$mathrm L(x)=logleft[left(fracpieright)^xj(x)right]+sum_ngeq1log E_n(x)$$
Then we define $$U(x)=prod_ngeq1E_n(x)$$
To see that $$U(x)=left(fracepi xright)^xexpmathrm L(x)$$
Where we used $$sum_nlog(a_n)=logleft[prod_na_nright]$$
and the neat rules $$log(a^b)=log(e^blog a)=blog a$$
$$log(a)pm b=logleft(e^pm baright)$$
to simplify the expressions. Next, we define
$$P_mu,nu(a_1,a_2,dots,a_mu;b_1,b_2,dots,b_nu)=fracprod_i=1^mu U(a_i)prod_i=1^nu U(b_i)$$
And we see that
$$P_mu,nu(a_1,dots,a_mu;b_1,dots,b_nu)=prod_ngeq1fracprod_i=1^mu E_n(a_i)prod_i=1^nu E_n(b_i)$$
This gives $$P_1,1(x_1;x_2)=left(fracepiright)^x_1-x_2fracj(x_2)j(x_1)expleft[mathrm L(x_1)-mathrm L(x_2)right]$$
Then we define $$mathrmT(x)=frac1piint_0^pi xlog(tan t)dt=mathrm L(x)-mathrm L(x+1/2)-frac12log2$$
To get that
$$P_1,1left(x;x+frac12right)=sqrtfrac2pie,fracj(x+1/2)j(x)expmathrm T(x)$$
So we have
$$P_2,2left(x_1,x_2+frac12 ;x_2,x_1+frac12right)=fracj(x_1+1/2)j(x_2)j(x_2+1/2)j(x_1)expleft[mathrm T(x_1)-mathrm T(x_2)right]$$
Then using the identities
$$mathrm L(1/2)=-frac12log2$$
$$mathrm L(1/4)=fracmathrm G2pi-frac14log2$$
We get $$P_1,1left(frac12;frac14right)=frac1(2pi)^1/4expleft[fracmathrm G2pi+frac14right]tag1$$
From here, the identity
$$-mathrm T(1/12)=frac2mathrm G3pi$$
which gives
$$P_1,1left(frac712;frac112right)=sqrtfrac67pisqrt[6]7expleft[frac2mathrm G3pi+frac12right]tag2$$
Then from here, the identity
$$mathrm T(1/20)-mathrm T(3/20)=frac2mathrm G5pi$$
gives $$P_2,2left(frac120,frac1320;frac320,frac1120right)=left(fracj(11)j(3)j(13)right)^1/20expfrac2mathrm G5pitag3$$
Then multiplying $(1),(2),$ and $(3)$, we have the desired result, namely
$$P_4,4left(frac12,frac712,frac120,frac1320;frac14,frac112,frac320,frac1120right)=alpha$$
calculus integration alternative-proof infinite-product constants
asked Mar 13 at 17:43
clathratusclathratus
5,0601338
$endgroup$
$begingroup$
Were you ask a question? or just write a proof?
$endgroup$
– BarzanHayati
Mar 13 at 21:26
$begingroup$
See the edit. My question is highlighted in yellow
$endgroup$
– clathratus
Mar 13 at 21:33
add a comment |
$begingroup$
$mathrm G$ is Catalan's constant.
I recently found the product
$$
alpha=prod_n=1^inftyfracE_n(frac12)E_n(frac712)E_n(frac120)E_n(frac1320)E_n(frac14)E_n(frac112)E_n(frac320)E_n(frac1120)=\
expleft[frac47mathrm G30pi+frac34right]sqrtfrac3391pisqrtfrac2pifracsqrt[5]11sqrt[3]7sqrt[5]frac3^313^3$$
(an alternate form of the product in the title)
Where $$E_n(x)=fracj(n+x)(en)^2xj(n-x)qquad xin(0,1)$$
and $j(x)=x^x$.
Could I have some numerical evidence, or better yet an alternate proof? My tools are limited to desmos, which cannot really handle infinite products. Thanks.
My Proof.
We define $$mathrm L(x)=frac1piint_0^pi xlog(sin t)dt$$
And we use $$sin t=tprod_ngeq1left(1-fract^2pi^2 n^2right)$$
To see that $$log(sin t)=log(t)+sum_ngeq1logfracpi^2n^2-t^2pi^2n^2$$
Then integrate both sides over $[0,x]$ to get
$$pimathrm L(x/pi)=x(log x-1)+sum_ngeq1xlogbigg(1-fracx^2pi^2n^2bigg)-2x+pi nlogfracpi n+xpi n-x$$
$$pimathrm L(x/pi)=logleft[fracj(x)e^xright]+sum_ngeq1logleft[fracj(pi n+x)(epi n)^2xj(pi n-x)right]$$
$xmapsto pi x$:
$$pimathrm L(x)=logleft[fracj(pi x)e^pi xright]+sum_ngeq1logleft[fracj(pi n+pi x)(epi n)^2pi xj(pi n-pi x)right]$$
$$mathrm L(x)=logleft[left(fracpieright)^xj(x)right]+sum_ngeq1log E_n(x)$$
Then we define $$U(x)=prod_ngeq1E_n(x)$$
To see that $$U(x)=left(fracepi xright)^xexpmathrm L(x)$$
Where we used $$sum_nlog(a_n)=logleft[prod_na_nright]$$
and the neat rules $$log(a^b)=log(e^blog a)=blog a$$
$$log(a)pm b=logleft(e^pm baright)$$
to simplify the expressions. Next, we define
$$P_mu,nu(a_1,a_2,dots,a_mu;b_1,b_2,dots,b_nu)=fracprod_i=1^mu U(a_i)prod_i=1^nu U(b_i)$$
And we see that
$$P_mu,nu(a_1,dots,a_mu;b_1,dots,b_nu)=prod_ngeq1fracprod_i=1^mu E_n(a_i)prod_i=1^nu E_n(b_i)$$
This gives $$P_1,1(x_1;x_2)=left(fracepiright)^x_1-x_2fracj(x_2)j(x_1)expleft[mathrm L(x_1)-mathrm L(x_2)right]$$
Then we define $$mathrmT(x)=frac1piint_0^pi xlog(tan t)dt=mathrm L(x)-mathrm L(x+1/2)-frac12log2$$
To get that
$$P_1,1left(x;x+frac12right)=sqrtfrac2pie,fracj(x+1/2)j(x)expmathrm T(x)$$
So we have
$$P_2,2left(x_1,x_2+frac12 ;x_2,x_1+frac12right)=fracj(x_1+1/2)j(x_2)j(x_2+1/2)j(x_1)expleft[mathrm T(x_1)-mathrm T(x_2)right]$$
Then using the identities
$$mathrm L(1/2)=-frac12log2$$
$$mathrm L(1/4)=fracmathrm G2pi-frac14log2$$
We get $$P_1,1left(frac12;frac14right)=frac1(2pi)^1/4expleft[fracmathrm G2pi+frac14right]tag1$$
From here, the identity
$$-mathrm T(1/12)=frac2mathrm G3pi$$
which gives
$$P_1,1left(frac712;frac112right)=sqrtfrac67pisqrt[6]7expleft[frac2mathrm G3pi+frac12right]tag2$$
Then from here, the identity
$$mathrm T(1/20)-mathrm T(3/20)=frac2mathrm G5pi$$
gives $$P_2,2left(frac120,frac1320;frac320,frac1120right)=left(fracj(11)j(3)j(13)right)^1/20expfrac2mathrm G5pitag3$$
Then multiplying $(1),(2),$ and $(3)$, we have the desired result, namely
$$P_4,4left(frac12,frac712,frac120,frac1320;frac14,frac112,frac320,frac1120right)=alpha$$
calculus integration alternative-proof infinite-product constants
asked Mar 13 at 17:43
clathratusclathratus
5,0601338
$endgroup$
$mathrm G$ is Catalan's constant.
I recently found the product
$$
alpha=prod_n=1^inftyfracE_n(frac12)E_n(frac712)E_n(frac120)E_n(frac1320)E_n(frac14)E_n(frac112)E_n(frac320)E_n(frac1120)=\
expleft[frac47mathrm G30pi+frac34right]sqrtfrac3391pisqrtfrac2pifracsqrt[5]11sqrt[3]7sqrt[5]frac3^313^3$$
(an alternate form of the product in the title)
Where $$E_n(x)=fracj(n+x)(en)^2xj(n-x)qquad xin(0,1)$$
and $j(x)=x^x$.
Could I have some numerical evidence, or better yet an alternate proof? My tools are limited to desmos, which cannot really handle infinite products. Thanks.
My Proof.
We define $$mathrm L(x)=frac1piint_0^pi xlog(sin t)dt$$
And we use $$sin t=tprod_ngeq1left(1-fract^2pi^2 n^2right)$$
To see that $$log(sin t)=log(t)+sum_ngeq1logfracpi^2n^2-t^2pi^2n^2$$
Then integrate both sides over $[0,x]$ to get
$$pimathrm L(x/pi)=x(log x-1)+sum_ngeq1xlogbigg(1-fracx^2pi^2n^2bigg)-2x+pi nlogfracpi n+xpi n-x$$
$$pimathrm L(x/pi)=logleft[fracj(x)e^xright]+sum_ngeq1logleft[fracj(pi n+x)(epi n)^2xj(pi n-x)right]$$
$xmapsto pi x$:
$$pimathrm L(x)=logleft[fracj(pi x)e^pi xright]+sum_ngeq1logleft[fracj(pi n+pi x)(epi n)^2pi xj(pi n-pi x)right]$$
$$mathrm L(x)=logleft[left(fracpieright)^xj(x)right]+sum_ngeq1log E_n(x)$$
Then we define $$U(x)=prod_ngeq1E_n(x)$$
To see that $$U(x)=left(fracepi xright)^xexpmathrm L(x)$$
Where we used $$sum_nlog(a_n)=logleft[prod_na_nright]$$
and the neat rules $$log(a^b)=log(e^blog a)=blog a$$
$$log(a)pm b=logleft(e^pm baright)$$
to simplify the expressions. Next, we define
$$P_mu,nu(a_1,a_2,dots,a_mu;b_1,b_2,dots,b_nu)=fracprod_i=1^mu U(a_i)prod_i=1^nu U(b_i)$$
And we see that
$$P_mu,nu(a_1,dots,a_mu;b_1,dots,b_nu)=prod_ngeq1fracprod_i=1^mu E_n(a_i)prod_i=1^nu E_n(b_i)$$
This gives $$P_1,1(x_1;x_2)=left(fracepiright)^x_1-x_2fracj(x_2)j(x_1)expleft[mathrm L(x_1)-mathrm L(x_2)right]$$
Then we define $$mathrmT(x)=frac1piint_0^pi xlog(tan t)dt=mathrm L(x)-mathrm L(x+1/2)-frac12log2$$
To get that
$$P_1,1left(x;x+frac12right)=sqrtfrac2pie,fracj(x+1/2)j(x)expmathrm T(x)$$
So we have
$$P_2,2left(x_1,x_2+frac12 ;x_2,x_1+frac12right)=fracj(x_1+1/2)j(x_2)j(x_2+1/2)j(x_1)expleft[mathrm T(x_1)-mathrm T(x_2)right]$$
Then using the identities
$$mathrm L(1/2)=-frac12log2$$
$$mathrm L(1/4)=fracmathrm G2pi-frac14log2$$
We get $$P_1,1left(frac12;frac14right)=frac1(2pi)^1/4expleft[fracmathrm G2pi+frac14right]tag1$$
From here, the identity
$$-mathrm T(1/12)=frac2mathrm G3pi$$
which gives
$$P_1,1left(frac712;frac112right)=sqrtfrac67pisqrt[6]7expleft[frac2mathrm G3pi+frac12right]tag2$$
Then from here, the identity
$$mathrm T(1/20)-mathrm T(3/20)=frac2mathrm G5pi$$
gives $$P_2,2left(frac120,frac1320;frac320,frac1120right)=left(fracj(11)j(3)j(13)right)^1/20expfrac2mathrm G5pitag3$$
Then multiplying $(1),(2),$ and $(3)$, we have the desired result, namely
$$P_4,4left(frac12,frac712,frac120,frac1320;frac14,frac112,frac320,frac1120right)=alpha$$
calculus integration alternative-proof infinite-product constants
calculus integration alternative-proof infinite-product constants
asked Mar 13 at 17:43
clathratusclathratus
5,0601338
asked Mar 13 at 17:43
clathratusclathratus
5,0601338
edited 2 days ago
clathratus
asked Mar 13 at 17:43
clathratusclathratus
5,0601338
asked Mar 13 at 17:43
clathratusclathratus
5,0601338
asked Mar 13 at 17:43
clathratusclathratus
5,0601338
5,0601338
$begingroup$
Were you ask a question? or just write a proof?
$endgroup$
– BarzanHayati
Mar 13 at 21:26
$begingroup$
See the edit. My question is highlighted in yellow
$endgroup$
– clathratus
Mar 13 at 21:33
add a comment |
$begingroup$
Were you ask a question? or just write a proof?
$endgroup$
– BarzanHayati
Mar 13 at 21:26
$begingroup$
See the edit. My question is highlighted in yellow
$endgroup$
– clathratus
Mar 13 at 21:33
$begingroup$
Were you ask a question? or just write a proof?
$endgroup$
– BarzanHayati
Mar 13 at 21:26
$begingroup$
Were you ask a question? or just write a proof?
$endgroup$
– BarzanHayati
Mar 13 at 21:26
$begingroup$
See the edit. My question is highlighted in yellow
$endgroup$
– clathratus
Mar 13 at 21:33
$begingroup$
See the edit. My question is highlighted in yellow
$endgroup$
– clathratus
Mar 13 at 21:33
add a comment |
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$begingroup$
Were you ask a question? or just write a proof?
$endgroup$
– BarzanHayati
Mar 13 at 21:26
$begingroup$
See the edit. My question is highlighted in yellow
$endgroup$
– clathratus
Mar 13 at 21:33