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Is there another way of evaluating $lim_x to 0 Gamma(x)(gamma+psi(1+x))=fracpi^26$
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I was messing around with the Zeta-Function and I got what I thought was an interesting limit:
$$lim_x to 0 Gamma(x)(gamma+psi(1+x)) = fracpi^26 $$
Where $Gamma$ is the gamma function, $gamma $ is the Euler-Mascheroni constant, and $psi$ is the digamma function.
I got it by writing the Zeta Function as:
$$sum_n=1^infty frac1n^2 = -int_0^1 fracln(x)1-xdx$$
Then using the beta function and differentiating with respect to x you get the limit.
What other ways can be used to evaluate the limit?
Here's how I got to my answer:
We know
$$sum_n=1^inftyfrac1n^2 = fracpi^26 $$
To convert the sum into an integral consider:
$$int_0^1 x^n-1dx=frac1n$$
Differentiating once:
$$int_0^1 ln(x)x^n-1dx=frac-1n^2$$
Plugging in the integral into the sum you get:
$$-sum_n=1^infty int_0^1 ln(x)x^n-1dx=-int_0^1ln(x)sum_n=1^infty x^n-1dx=-int_0^1 fracln(x)1-xdx$$
Euler's Beta Function is defined as:
$$operatornameB(x,y)=int_0^1 t^x-1(1-t)^y-1 = fracGamma(x)Gamma(y)Gamma(x+y)$$
Differentiation with respect to x:
$$operatornameB_x(x,y)=int_0^1 ln(t)t^x-1(1-t)^y-1 = fracGamma(x)Gamma(y)Gamma(x+y)(psi(x)-psi(x+y))$$
$$-int_0^1 fracln(x)1-xdx = -operatornameB_x(1,0)=fracpi^26 $$
Taking the limit as (x,y)->(1,0)
$$-lim_(x,y)to(1,0) fracGamma(x)Gamma(y)Gamma(x+y)(psi(x)-psi(x+y))=-lim_y to 0 Gamma(y)(psi(1)-psi(1+y)) = lim_y to 0 Gamma(y)(gamma+psi(1+y))$$
integration limits summation gamma-function riemann-zeta
asked Aug 2 at 14:43
Tom Himler
547213
 |Â
show 1 more comment
up vote
2
down vote
favorite
I was messing around with the Zeta-Function and I got what I thought was an interesting limit:
$$lim_x to 0 Gamma(x)(gamma+psi(1+x)) = fracpi^26 $$
Where $Gamma$ is the gamma function, $gamma $ is the Euler-Mascheroni constant, and $psi$ is the digamma function.
I got it by writing the Zeta Function as:
$$sum_n=1^infty frac1n^2 = -int_0^1 fracln(x)1-xdx$$
Then using the beta function and differentiating with respect to x you get the limit.
What other ways can be used to evaluate the limit?
Here's how I got to my answer:
We know
$$sum_n=1^inftyfrac1n^2 = fracpi^26 $$
To convert the sum into an integral consider:
$$int_0^1 x^n-1dx=frac1n$$
Differentiating once:
$$int_0^1 ln(x)x^n-1dx=frac-1n^2$$
Plugging in the integral into the sum you get:
$$-sum_n=1^infty int_0^1 ln(x)x^n-1dx=-int_0^1ln(x)sum_n=1^infty x^n-1dx=-int_0^1 fracln(x)1-xdx$$
Euler's Beta Function is defined as:
$$operatornameB(x,y)=int_0^1 t^x-1(1-t)^y-1 = fracGamma(x)Gamma(y)Gamma(x+y)$$
Differentiation with respect to x:
$$operatornameB_x(x,y)=int_0^1 ln(t)t^x-1(1-t)^y-1 = fracGamma(x)Gamma(y)Gamma(x+y)(psi(x)-psi(x+y))$$
$$-int_0^1 fracln(x)1-xdx = -operatornameB_x(1,0)=fracpi^26 $$
Taking the limit as (x,y)->(1,0)
$$-lim_(x,y)to(1,0) fracGamma(x)Gamma(y)Gamma(x+y)(psi(x)-psi(x+y))=-lim_y to 0 Gamma(y)(psi(1)-psi(1+y)) = lim_y to 0 Gamma(y)(gamma+psi(1+y))$$
integration limits summation gamma-function riemann-zeta
asked Aug 2 at 14:43
Tom Himler
547213
1
What are $psi$ and $gamma$ though? Are these standard notations?
– Suzet
Aug 2 at 14:48
1
They're the digamma function and the euler-mascheroni constant. I'll edit the post so it's more clear.
– Tom Himler
Aug 2 at 14:49
1
I am just interested. How exactly did you derived the limit from the given integral representation with your steps. I tried to do it by myself and failed. Could you maybe explain it to me or add it to the question?
– mrtaurho
Aug 2 at 15:27
@mrtaurho I put how I got my answer if that helps. It's not super rigorous since I don't really know that sort of shtuff yet.
– Tom Himler
Aug 2 at 18:25
1
Thank you for adding it nevertheless. I am not that familiar as I want to with these kinds of reshapings and this stuff. It is great for me to learn more about it.
– mrtaurho
Aug 2 at 18:27
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I was messing around with the Zeta-Function and I got what I thought was an interesting limit:
$$lim_x to 0 Gamma(x)(gamma+psi(1+x)) = fracpi^26 $$
Where $Gamma$ is the gamma function, $gamma $ is the Euler-Mascheroni constant, and $psi$ is the digamma function.
I got it by writing the Zeta Function as:
$$sum_n=1^infty frac1n^2 = -int_0^1 fracln(x)1-xdx$$
Then using the beta function and differentiating with respect to x you get the limit.
What other ways can be used to evaluate the limit?
Here's how I got to my answer:
We know
$$sum_n=1^inftyfrac1n^2 = fracpi^26 $$
To convert the sum into an integral consider:
$$int_0^1 x^n-1dx=frac1n$$
Differentiating once:
$$int_0^1 ln(x)x^n-1dx=frac-1n^2$$
Plugging in the integral into the sum you get:
$$-sum_n=1^infty int_0^1 ln(x)x^n-1dx=-int_0^1ln(x)sum_n=1^infty x^n-1dx=-int_0^1 fracln(x)1-xdx$$
Euler's Beta Function is defined as:
$$operatornameB(x,y)=int_0^1 t^x-1(1-t)^y-1 = fracGamma(x)Gamma(y)Gamma(x+y)$$
Differentiation with respect to x:
$$operatornameB_x(x,y)=int_0^1 ln(t)t^x-1(1-t)^y-1 = fracGamma(x)Gamma(y)Gamma(x+y)(psi(x)-psi(x+y))$$
$$-int_0^1 fracln(x)1-xdx = -operatornameB_x(1,0)=fracpi^26 $$
Taking the limit as (x,y)->(1,0)
$$-lim_(x,y)to(1,0) fracGamma(x)Gamma(y)Gamma(x+y)(psi(x)-psi(x+y))=-lim_y to 0 Gamma(y)(psi(1)-psi(1+y)) = lim_y to 0 Gamma(y)(gamma+psi(1+y))$$
integration limits summation gamma-function riemann-zeta
asked Aug 2 at 14:43
Tom Himler
547213
I was messing around with the Zeta-Function and I got what I thought was an interesting limit:
$$lim_x to 0 Gamma(x)(gamma+psi(1+x)) = fracpi^26 $$
Where $Gamma$ is the gamma function, $gamma $ is the Euler-Mascheroni constant, and $psi$ is the digamma function.
I got it by writing the Zeta Function as:
$$sum_n=1^infty frac1n^2 = -int_0^1 fracln(x)1-xdx$$
Then using the beta function and differentiating with respect to x you get the limit.
What other ways can be used to evaluate the limit?
Here's how I got to my answer:
We know
$$sum_n=1^inftyfrac1n^2 = fracpi^26 $$
To convert the sum into an integral consider:
$$int_0^1 x^n-1dx=frac1n$$
Differentiating once:
$$int_0^1 ln(x)x^n-1dx=frac-1n^2$$
Plugging in the integral into the sum you get:
$$-sum_n=1^infty int_0^1 ln(x)x^n-1dx=-int_0^1ln(x)sum_n=1^infty x^n-1dx=-int_0^1 fracln(x)1-xdx$$
Euler's Beta Function is defined as:
$$operatornameB(x,y)=int_0^1 t^x-1(1-t)^y-1 = fracGamma(x)Gamma(y)Gamma(x+y)$$
Differentiation with respect to x:
$$operatornameB_x(x,y)=int_0^1 ln(t)t^x-1(1-t)^y-1 = fracGamma(x)Gamma(y)Gamma(x+y)(psi(x)-psi(x+y))$$
$$-int_0^1 fracln(x)1-xdx = -operatornameB_x(1,0)=fracpi^26 $$
Taking the limit as (x,y)->(1,0)
$$-lim_(x,y)to(1,0) fracGamma(x)Gamma(y)Gamma(x+y)(psi(x)-psi(x+y))=-lim_y to 0 Gamma(y)(psi(1)-psi(1+y)) = lim_y to 0 Gamma(y)(gamma+psi(1+y))$$
integration limits summation gamma-function riemann-zeta
asked Aug 2 at 14:43
Tom Himler
547213
edited Aug 2 at 18:24
asked Aug 2 at 14:43
Tom Himler
547213
asked Aug 2 at 14:43
Tom Himler
547213
asked Aug 2 at 14:43
Tom Himler
547213
547213
1
What are $psi$ and $gamma$ though? Are these standard notations?
– Suzet
Aug 2 at 14:48
1
They're the digamma function and the euler-mascheroni constant. I'll edit the post so it's more clear.
– Tom Himler
Aug 2 at 14:49
1
I am just interested. How exactly did you derived the limit from the given integral representation with your steps. I tried to do it by myself and failed. Could you maybe explain it to me or add it to the question?
– mrtaurho
Aug 2 at 15:27
@mrtaurho I put how I got my answer if that helps. It's not super rigorous since I don't really know that sort of shtuff yet.
– Tom Himler
Aug 2 at 18:25
1
Thank you for adding it nevertheless. I am not that familiar as I want to with these kinds of reshapings and this stuff. It is great for me to learn more about it.
– mrtaurho
Aug 2 at 18:27
 |Â
show 1 more comment
1
What are $psi$ and $gamma$ though? Are these standard notations?
– Suzet
Aug 2 at 14:48
1
They're the digamma function and the euler-mascheroni constant. I'll edit the post so it's more clear.
– Tom Himler
Aug 2 at 14:49
1
I am just interested. How exactly did you derived the limit from the given integral representation with your steps. I tried to do it by myself and failed. Could you maybe explain it to me or add it to the question?
– mrtaurho
Aug 2 at 15:27
@mrtaurho I put how I got my answer if that helps. It's not super rigorous since I don't really know that sort of shtuff yet.
– Tom Himler
Aug 2 at 18:25
1
Thank you for adding it nevertheless. I am not that familiar as I want to with these kinds of reshapings and this stuff. It is great for me to learn more about it.
– mrtaurho
Aug 2 at 18:27
1
1
What are $psi$ and $gamma$ though? Are these standard notations?
– Suzet
Aug 2 at 14:48
What are $psi$ and $gamma$ though? Are these standard notations?
– Suzet
Aug 2 at 14:48
1
1
They're the digamma function and the euler-mascheroni constant. I'll edit the post so it's more clear.
– Tom Himler
Aug 2 at 14:49
They're the digamma function and the euler-mascheroni constant. I'll edit the post so it's more clear.
– Tom Himler
Aug 2 at 14:49
1
1
I am just interested. How exactly did you derived the limit from the given integral representation with your steps. I tried to do it by myself and failed. Could you maybe explain it to me or add it to the question?
– mrtaurho
Aug 2 at 15:27
I am just interested. How exactly did you derived the limit from the given integral representation with your steps. I tried to do it by myself and failed. Could you maybe explain it to me or add it to the question?
– mrtaurho
Aug 2 at 15:27
@mrtaurho I put how I got my answer if that helps. It's not super rigorous since I don't really know that sort of shtuff yet.
– Tom Himler
Aug 2 at 18:25
@mrtaurho I put how I got my answer if that helps. It's not super rigorous since I don't really know that sort of shtuff yet.
– Tom Himler
Aug 2 at 18:25
1
1
Thank you for adding it nevertheless. I am not that familiar as I want to with these kinds of reshapings and this stuff. It is great for me to learn more about it.
– mrtaurho
Aug 2 at 18:27
Thank you for adding it nevertheless. I am not that familiar as I want to with these kinds of reshapings and this stuff. It is great for me to learn more about it.
– mrtaurho
Aug 2 at 18:27
 |Â
show 1 more comment
1 Answer
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up vote
7
down vote
One may use $Gamma(x)=Gamma(x+1)/x$ to get
$$lim_xto0fracgamma+psi(1+x)x$$
and since $psi(1)=-gamma$, we may use the definition of the derivative to get $psi'(1)$, which has many different forms, of which the result may be deduced from.
answered Aug 2 at 14:55
Simply Beautiful Art
49.1k571169
Thanks! That's very clever.
– Tom Himler
Aug 2 at 14:58
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
One may use $Gamma(x)=Gamma(x+1)/x$ to get
$$lim_xto0fracgamma+psi(1+x)x$$
and since $psi(1)=-gamma$, we may use the definition of the derivative to get $psi'(1)$, which has many different forms, of which the result may be deduced from.
answered Aug 2 at 14:55
Simply Beautiful Art
49.1k571169
Thanks! That's very clever.
– Tom Himler
Aug 2 at 14:58
add a comment |Â
up vote
7
down vote
One may use $Gamma(x)=Gamma(x+1)/x$ to get
$$lim_xto0fracgamma+psi(1+x)x$$
and since $psi(1)=-gamma$, we may use the definition of the derivative to get $psi'(1)$, which has many different forms, of which the result may be deduced from.
answered Aug 2 at 14:55
Simply Beautiful Art
49.1k571169
Thanks! That's very clever.
– Tom Himler
Aug 2 at 14:58
add a comment |Â
up vote
7
down vote
up vote
7
down vote
One may use $Gamma(x)=Gamma(x+1)/x$ to get
$$lim_xto0fracgamma+psi(1+x)x$$
and since $psi(1)=-gamma$, we may use the definition of the derivative to get $psi'(1)$, which has many different forms, of which the result may be deduced from.
answered Aug 2 at 14:55
Simply Beautiful Art
49.1k571169
One may use $Gamma(x)=Gamma(x+1)/x$ to get
$$lim_xto0fracgamma+psi(1+x)x$$
and since $psi(1)=-gamma$, we may use the definition of the derivative to get $psi'(1)$, which has many different forms, of which the result may be deduced from.
answered Aug 2 at 14:55
Simply Beautiful Art
49.1k571169
answered Aug 2 at 14:55
Simply Beautiful Art
49.1k571169
answered Aug 2 at 14:55
Simply Beautiful Art
49.1k571169
answered Aug 2 at 14:55
Simply Beautiful Art
49.1k571169
49.1k571169
Thanks! That's very clever.
– Tom Himler
Aug 2 at 14:58
add a comment |Â
Thanks! That's very clever.
– Tom Himler
Aug 2 at 14:58
Thanks! That's very clever.
– Tom Himler
Aug 2 at 14:58
Thanks! That's very clever.
– Tom Himler
Aug 2 at 14:58
add a comment |Â
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1
What are $psi$ and $gamma$ though? Are these standard notations?
– Suzet
Aug 2 at 14:48
1
They're the digamma function and the euler-mascheroni constant. I'll edit the post so it's more clear.
– Tom Himler
Aug 2 at 14:49
1
I am just interested. How exactly did you derived the limit from the given integral representation with your steps. I tried to do it by myself and failed. Could you maybe explain it to me or add it to the question?
– mrtaurho
Aug 2 at 15:27
@mrtaurho I put how I got my answer if that helps. It's not super rigorous since I don't really know that sort of shtuff yet.
– Tom Himler
Aug 2 at 18:25
1
Thank you for adding it nevertheless. I am not that familiar as I want to with these kinds of reshapings and this stuff. It is great for me to learn more about it.
– mrtaurho
Aug 2 at 18:27