Mixing
Definite Integral over the Gamma Function
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Do we have any methods for evaluating $$int_1^infty frac1Gamma(s) ,ds$$?
I thought about perhaps rewriting as $$int_1^infty fracGamma(1-s)Gamma(1-s) Gamma(s) ,ds$$
$$=frac1pi int_1^infty Gamma(1-s) sin(pi s) ,ds $$
But I'm not too sure if this is all that useful. Thoughts?
integration definite-integrals gamma-function
asked Sep 4 at 1:49
Ryan Goulden
338110
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up vote
3
down vote
favorite
Do we have any methods for evaluating $$int_1^infty frac1Gamma(s) ,ds$$?
I thought about perhaps rewriting as $$int_1^infty fracGamma(1-s)Gamma(1-s) Gamma(s) ,ds$$
$$=frac1pi int_1^infty Gamma(1-s) sin(pi s) ,ds $$
But I'm not too sure if this is all that useful. Thoughts?
integration definite-integrals gamma-function
asked Sep 4 at 1:49
Ryan Goulden
338110
1
Maybe use the integral definition for the gamma function on the last line? I mean, what makes you so certain that a closed form exists?
– Frank W.
Sep 4 at 2:13
en.wikipedia.org/wiki/Frans%C3%A9n%E2%80%93Robinson_constant
– Harmonic Sun
Sep 4 at 2:24
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Do we have any methods for evaluating $$int_1^infty frac1Gamma(s) ,ds$$?
I thought about perhaps rewriting as $$int_1^infty fracGamma(1-s)Gamma(1-s) Gamma(s) ,ds$$
$$=frac1pi int_1^infty Gamma(1-s) sin(pi s) ,ds $$
But I'm not too sure if this is all that useful. Thoughts?
integration definite-integrals gamma-function
asked Sep 4 at 1:49
Ryan Goulden
338110
Do we have any methods for evaluating $$int_1^infty frac1Gamma(s) ,ds$$?
I thought about perhaps rewriting as $$int_1^infty fracGamma(1-s)Gamma(1-s) Gamma(s) ,ds$$
$$=frac1pi int_1^infty Gamma(1-s) sin(pi s) ,ds $$
But I'm not too sure if this is all that useful. Thoughts?
integration definite-integrals gamma-function
asked Sep 4 at 1:49
Ryan Goulden
338110
asked Sep 4 at 1:49
Ryan Goulden
338110
asked Sep 4 at 1:49
Ryan Goulden
338110
asked Sep 4 at 1:49
Ryan Goulden
338110
338110
1
Maybe use the integral definition for the gamma function on the last line? I mean, what makes you so certain that a closed form exists?
– Frank W.
Sep 4 at 2:13
en.wikipedia.org/wiki/Frans%C3%A9n%E2%80%93Robinson_constant
– Harmonic Sun
Sep 4 at 2:24
add a comment |Â
1
Maybe use the integral definition for the gamma function on the last line? I mean, what makes you so certain that a closed form exists?
– Frank W.
Sep 4 at 2:13
en.wikipedia.org/wiki/Frans%C3%A9n%E2%80%93Robinson_constant
– Harmonic Sun
Sep 4 at 2:24
1
1
Maybe use the integral definition for the gamma function on the last line? I mean, what makes you so certain that a closed form exists?
– Frank W.
Sep 4 at 2:13
Maybe use the integral definition for the gamma function on the last line? I mean, what makes you so certain that a closed form exists?
– Frank W.
Sep 4 at 2:13
en.wikipedia.org/wiki/Frans%C3%A9n%E2%80%93Robinson_constant
– Harmonic Sun
Sep 4 at 2:24
en.wikipedia.org/wiki/Frans%C3%A9n%E2%80%93Robinson_constant
– Harmonic Sun
Sep 4 at 2:24
add a comment |Â
2 Answers
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$$frac1piint_1^inftyGamma(1-s)sin(pi s)ds=frac1piint_0^infty e^-xint_1^infty sin(pi s)e^-slog(x)dsdx=-frac1piint_0^infty e^-xfrac fracpixlog^2(x)+pi^2dx=-int_0^infty frace^-xxfracdxlog^2(x)+pi^2$$ which looks similar to the Fransen-Robinson Constant and has no known closed form.
answered Sep 4 at 2:36
aleden
1,277311
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up vote
3
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If we want to evaluate this integral, series expansion are not too bad.
Around $s=0$
$$ frac1Gamma(s)=s+gamma s^2+left(fracgamma ^22-fracpi ^212right) s^3+frac112
s^4 left(2 gamma ^3-gamma pi ^2-2 psi ^(2)(1)right)+Oleft(s^5right)tag 1$$
Around $s=1$
$$ frac1Gamma(s)=1+gamma (s-1)+left(fracgamma ^22-fracpi ^212right)
(s-1)^2+frac112 (s-1)^3 left(2 gamma ^3-gamma pi ^2-2 psi
^(2)(1)right)+frac(s-1)^4 left(60 gamma ^4-60 gamma ^2 pi ^2+pi ^4-240
gamma psi ^(2)(1)right)1440+Oleft((s-1)^5right)tag 2$$
As a function of the order of the expansions, we would get (in decimal representation)
$$left(
beginarrayccc
n & (1) & (2) \
1 & 0.500000 & 0.711392 \
2 & 0.692405 & 0.492766 \
3 & 0.528436 & 0.503267 \
4 & 0.520035 & 0.536575 \
5 & 0.547792 & 0.543607 \
6 & 0.541763 & 0.542233 \
7 & 0.540561 & 0.541331 \
8 & 0.541363 & 0.541201 \
9 & 0.541246 & 0.541223 \
10 & 0.541227 & 0.541234
endarray
right)$$ for an exact value $approx 0.541236$.
answered Sep 4 at 4:56
Claude Leibovici
113k1155127
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
$$frac1piint_1^inftyGamma(1-s)sin(pi s)ds=frac1piint_0^infty e^-xint_1^infty sin(pi s)e^-slog(x)dsdx=-frac1piint_0^infty e^-xfrac fracpixlog^2(x)+pi^2dx=-int_0^infty frace^-xxfracdxlog^2(x)+pi^2$$ which looks similar to the Fransen-Robinson Constant and has no known closed form.
answered Sep 4 at 2:36
aleden
1,277311
add a comment |Â
up vote
5
down vote
$$frac1piint_1^inftyGamma(1-s)sin(pi s)ds=frac1piint_0^infty e^-xint_1^infty sin(pi s)e^-slog(x)dsdx=-frac1piint_0^infty e^-xfrac fracpixlog^2(x)+pi^2dx=-int_0^infty frace^-xxfracdxlog^2(x)+pi^2$$ which looks similar to the Fransen-Robinson Constant and has no known closed form.
answered Sep 4 at 2:36
aleden
1,277311
add a comment |Â
up vote
5
down vote
up vote
5
down vote
$$frac1piint_1^inftyGamma(1-s)sin(pi s)ds=frac1piint_0^infty e^-xint_1^infty sin(pi s)e^-slog(x)dsdx=-frac1piint_0^infty e^-xfrac fracpixlog^2(x)+pi^2dx=-int_0^infty frace^-xxfracdxlog^2(x)+pi^2$$ which looks similar to the Fransen-Robinson Constant and has no known closed form.
answered Sep 4 at 2:36
aleden
1,277311
$$frac1piint_1^inftyGamma(1-s)sin(pi s)ds=frac1piint_0^infty e^-xint_1^infty sin(pi s)e^-slog(x)dsdx=-frac1piint_0^infty e^-xfrac fracpixlog^2(x)+pi^2dx=-int_0^infty frace^-xxfracdxlog^2(x)+pi^2$$ which looks similar to the Fransen-Robinson Constant and has no known closed form.
answered Sep 4 at 2:36
aleden
1,277311
answered Sep 4 at 2:36
aleden
1,277311
answered Sep 4 at 2:36
aleden
1,277311
answered Sep 4 at 2:36
aleden
1,277311
1,277311
add a comment |Â
add a comment |Â
up vote
3
down vote
If we want to evaluate this integral, series expansion are not too bad.
Around $s=0$
$$ frac1Gamma(s)=s+gamma s^2+left(fracgamma ^22-fracpi ^212right) s^3+frac112
s^4 left(2 gamma ^3-gamma pi ^2-2 psi ^(2)(1)right)+Oleft(s^5right)tag 1$$
Around $s=1$
$$ frac1Gamma(s)=1+gamma (s-1)+left(fracgamma ^22-fracpi ^212right)
(s-1)^2+frac112 (s-1)^3 left(2 gamma ^3-gamma pi ^2-2 psi
^(2)(1)right)+frac(s-1)^4 left(60 gamma ^4-60 gamma ^2 pi ^2+pi ^4-240
gamma psi ^(2)(1)right)1440+Oleft((s-1)^5right)tag 2$$
As a function of the order of the expansions, we would get (in decimal representation)
$$left(
beginarrayccc
n & (1) & (2) \
1 & 0.500000 & 0.711392 \
2 & 0.692405 & 0.492766 \
3 & 0.528436 & 0.503267 \
4 & 0.520035 & 0.536575 \
5 & 0.547792 & 0.543607 \
6 & 0.541763 & 0.542233 \
7 & 0.540561 & 0.541331 \
8 & 0.541363 & 0.541201 \
9 & 0.541246 & 0.541223 \
10 & 0.541227 & 0.541234
endarray
right)$$ for an exact value $approx 0.541236$.
answered Sep 4 at 4:56
Claude Leibovici
113k1155127
add a comment |Â
up vote
3
down vote
If we want to evaluate this integral, series expansion are not too bad.
Around $s=0$
$$ frac1Gamma(s)=s+gamma s^2+left(fracgamma ^22-fracpi ^212right) s^3+frac112
s^4 left(2 gamma ^3-gamma pi ^2-2 psi ^(2)(1)right)+Oleft(s^5right)tag 1$$
Around $s=1$
$$ frac1Gamma(s)=1+gamma (s-1)+left(fracgamma ^22-fracpi ^212right)
(s-1)^2+frac112 (s-1)^3 left(2 gamma ^3-gamma pi ^2-2 psi
^(2)(1)right)+frac(s-1)^4 left(60 gamma ^4-60 gamma ^2 pi ^2+pi ^4-240
gamma psi ^(2)(1)right)1440+Oleft((s-1)^5right)tag 2$$
As a function of the order of the expansions, we would get (in decimal representation)
$$left(
beginarrayccc
n & (1) & (2) \
1 & 0.500000 & 0.711392 \
2 & 0.692405 & 0.492766 \
3 & 0.528436 & 0.503267 \
4 & 0.520035 & 0.536575 \
5 & 0.547792 & 0.543607 \
6 & 0.541763 & 0.542233 \
7 & 0.540561 & 0.541331 \
8 & 0.541363 & 0.541201 \
9 & 0.541246 & 0.541223 \
10 & 0.541227 & 0.541234
endarray
right)$$ for an exact value $approx 0.541236$.
answered Sep 4 at 4:56
Claude Leibovici
113k1155127
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If we want to evaluate this integral, series expansion are not too bad.
Around $s=0$
$$ frac1Gamma(s)=s+gamma s^2+left(fracgamma ^22-fracpi ^212right) s^3+frac112
s^4 left(2 gamma ^3-gamma pi ^2-2 psi ^(2)(1)right)+Oleft(s^5right)tag 1$$
Around $s=1$
$$ frac1Gamma(s)=1+gamma (s-1)+left(fracgamma ^22-fracpi ^212right)
(s-1)^2+frac112 (s-1)^3 left(2 gamma ^3-gamma pi ^2-2 psi
^(2)(1)right)+frac(s-1)^4 left(60 gamma ^4-60 gamma ^2 pi ^2+pi ^4-240
gamma psi ^(2)(1)right)1440+Oleft((s-1)^5right)tag 2$$
As a function of the order of the expansions, we would get (in decimal representation)
$$left(
beginarrayccc
n & (1) & (2) \
1 & 0.500000 & 0.711392 \
2 & 0.692405 & 0.492766 \
3 & 0.528436 & 0.503267 \
4 & 0.520035 & 0.536575 \
5 & 0.547792 & 0.543607 \
6 & 0.541763 & 0.542233 \
7 & 0.540561 & 0.541331 \
8 & 0.541363 & 0.541201 \
9 & 0.541246 & 0.541223 \
10 & 0.541227 & 0.541234
endarray
right)$$ for an exact value $approx 0.541236$.
answered Sep 4 at 4:56
Claude Leibovici
113k1155127
If we want to evaluate this integral, series expansion are not too bad.
Around $s=0$
$$ frac1Gamma(s)=s+gamma s^2+left(fracgamma ^22-fracpi ^212right) s^3+frac112
s^4 left(2 gamma ^3-gamma pi ^2-2 psi ^(2)(1)right)+Oleft(s^5right)tag 1$$
Around $s=1$
$$ frac1Gamma(s)=1+gamma (s-1)+left(fracgamma ^22-fracpi ^212right)
(s-1)^2+frac112 (s-1)^3 left(2 gamma ^3-gamma pi ^2-2 psi
^(2)(1)right)+frac(s-1)^4 left(60 gamma ^4-60 gamma ^2 pi ^2+pi ^4-240
gamma psi ^(2)(1)right)1440+Oleft((s-1)^5right)tag 2$$
As a function of the order of the expansions, we would get (in decimal representation)
$$left(
beginarrayccc
n & (1) & (2) \
1 & 0.500000 & 0.711392 \
2 & 0.692405 & 0.492766 \
3 & 0.528436 & 0.503267 \
4 & 0.520035 & 0.536575 \
5 & 0.547792 & 0.543607 \
6 & 0.541763 & 0.542233 \
7 & 0.540561 & 0.541331 \
8 & 0.541363 & 0.541201 \
9 & 0.541246 & 0.541223 \
10 & 0.541227 & 0.541234
endarray
right)$$ for an exact value $approx 0.541236$.
answered Sep 4 at 4:56
Claude Leibovici
113k1155127
answered Sep 4 at 4:56
Claude Leibovici
113k1155127
answered Sep 4 at 4:56
Claude Leibovici
113k1155127
answered Sep 4 at 4:56
Claude Leibovici
113k1155127
113k1155127
add a comment |Â
add a comment |Â
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1
Maybe use the integral definition for the gamma function on the last line? I mean, what makes you so certain that a closed form exists?
– Frank W.
Sep 4 at 2:13
en.wikipedia.org/wiki/Frans%C3%A9n%E2%80%93Robinson_constant
– Harmonic Sun
Sep 4 at 2:24