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Showing that $Gamma(x) Gamma(y) = Gamma(x+y)intlimits_0^1 lambda^x-1(1-lambda)^y-1dlambda$
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On page 56 of Titchmarsh's Theory of Functions, Titchmarsh makes the following claim:
beginalign
fracGamma(x)Gamma(y)Gamma(x+y) &= phi(x,y); (x>0,y>0)
endalign
where beginalign
phi(x,y) &=intlimits_0^inftyfracv^y-1(1+v)^x+ydv\
&= 2 intlimits_0^frac12pi (cos theta)^2x-1(sintheta)^2y-1dtheta\
&= intlimits_0^1 lambda^x-1(1-lambda)^y-1dlambda
endalign
I understand how the first integral is achieved using substitution, but I can't figure out how to go from the first integral to the second or the second integral to the third.
integration complex-analysis improper-integrals gamma-function
asked Jul 16 at 4:18
Josh
477
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On page 56 of Titchmarsh's Theory of Functions, Titchmarsh makes the following claim:
beginalign
fracGamma(x)Gamma(y)Gamma(x+y) &= phi(x,y); (x>0,y>0)
endalign
where beginalign
phi(x,y) &=intlimits_0^inftyfracv^y-1(1+v)^x+ydv\
&= 2 intlimits_0^frac12pi (cos theta)^2x-1(sintheta)^2y-1dtheta\
&= intlimits_0^1 lambda^x-1(1-lambda)^y-1dlambda
endalign
I understand how the first integral is achieved using substitution, but I can't figure out how to go from the first integral to the second or the second integral to the third.
integration complex-analysis improper-integrals gamma-function
asked Jul 16 at 4:18
Josh
477
Search for the derivation of the relation between beta and gamma functions.
– StubbornAtom
Jul 16 at 4:32
Second to third is by the substitution $lambda = (cos x)^2$
– Kavi Rama Murthy
Jul 16 at 5:55
See math.stackexchange.com/questions/514025/….
– StubbornAtom
Jul 16 at 6:15
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
On page 56 of Titchmarsh's Theory of Functions, Titchmarsh makes the following claim:
beginalign
fracGamma(x)Gamma(y)Gamma(x+y) &= phi(x,y); (x>0,y>0)
endalign
where beginalign
phi(x,y) &=intlimits_0^inftyfracv^y-1(1+v)^x+ydv\
&= 2 intlimits_0^frac12pi (cos theta)^2x-1(sintheta)^2y-1dtheta\
&= intlimits_0^1 lambda^x-1(1-lambda)^y-1dlambda
endalign
I understand how the first integral is achieved using substitution, but I can't figure out how to go from the first integral to the second or the second integral to the third.
integration complex-analysis improper-integrals gamma-function
asked Jul 16 at 4:18
Josh
477
On page 56 of Titchmarsh's Theory of Functions, Titchmarsh makes the following claim:
beginalign
fracGamma(x)Gamma(y)Gamma(x+y) &= phi(x,y); (x>0,y>0)
endalign
where beginalign
phi(x,y) &=intlimits_0^inftyfracv^y-1(1+v)^x+ydv\
&= 2 intlimits_0^frac12pi (cos theta)^2x-1(sintheta)^2y-1dtheta\
&= intlimits_0^1 lambda^x-1(1-lambda)^y-1dlambda
endalign
I understand how the first integral is achieved using substitution, but I can't figure out how to go from the first integral to the second or the second integral to the third.
integration complex-analysis improper-integrals gamma-function
asked Jul 16 at 4:18
Josh
477
asked Jul 16 at 4:18
Josh
477
asked Jul 16 at 4:18
Josh
477
asked Jul 16 at 4:18
Josh
477
477
Search for the derivation of the relation between beta and gamma functions.
– StubbornAtom
Jul 16 at 4:32
Second to third is by the substitution $lambda = (cos x)^2$
– Kavi Rama Murthy
Jul 16 at 5:55
See math.stackexchange.com/questions/514025/….
– StubbornAtom
Jul 16 at 6:15
add a comment |Â
Search for the derivation of the relation between beta and gamma functions.
– StubbornAtom
Jul 16 at 4:32
Second to third is by the substitution $lambda = (cos x)^2$
– Kavi Rama Murthy
Jul 16 at 5:55
See math.stackexchange.com/questions/514025/….
– StubbornAtom
Jul 16 at 6:15
Search for the derivation of the relation between beta and gamma functions.
– StubbornAtom
Jul 16 at 4:32
Search for the derivation of the relation between beta and gamma functions.
– StubbornAtom
Jul 16 at 4:32
Second to third is by the substitution $lambda = (cos x)^2$
– Kavi Rama Murthy
Jul 16 at 5:55
Second to third is by the substitution $lambda = (cos x)^2$
– Kavi Rama Murthy
Jul 16 at 5:55
See math.stackexchange.com/questions/514025/….
– StubbornAtom
Jul 16 at 6:15
See math.stackexchange.com/questions/514025/….
– StubbornAtom
Jul 16 at 6:15
add a comment |Â
1 Answer
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From the first to second,
beginalign
phi(x,y) &=intlimits_0^inftyfracv^y-1(1+v)^x+ydv\
&=intlimits_0^inftyfrac(tan^2theta)^y-1(1+(tan^2theta))^x+yd(tan^2theta)\
&=intlimits_0^frac12pifrac(tan^2theta)^y-1(sec^2theta)^x+y(2tanthetasec^2theta)dtheta\
&=2intlimits_0^frac12pi(sintheta)^2y-2+1(cos theta)^-(2y-2+1)+(2x+2y)-2dtheta\
&= 2 intlimits_0^frac12pi (cos theta)^2x-1(sintheta)^2y-1dtheta\
endalign
Now can you think of the next substitution? (Hint: look at the limits and the exponents)
answered Jul 16 at 4:36
Karn Watcharasupat
3,8172426
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
From the first to second,
beginalign
phi(x,y) &=intlimits_0^inftyfracv^y-1(1+v)^x+ydv\
&=intlimits_0^inftyfrac(tan^2theta)^y-1(1+(tan^2theta))^x+yd(tan^2theta)\
&=intlimits_0^frac12pifrac(tan^2theta)^y-1(sec^2theta)^x+y(2tanthetasec^2theta)dtheta\
&=2intlimits_0^frac12pi(sintheta)^2y-2+1(cos theta)^-(2y-2+1)+(2x+2y)-2dtheta\
&= 2 intlimits_0^frac12pi (cos theta)^2x-1(sintheta)^2y-1dtheta\
endalign
Now can you think of the next substitution? (Hint: look at the limits and the exponents)
answered Jul 16 at 4:36
Karn Watcharasupat
3,8172426
add a comment |Â
up vote
0
down vote
accepted
From the first to second,
beginalign
phi(x,y) &=intlimits_0^inftyfracv^y-1(1+v)^x+ydv\
&=intlimits_0^inftyfrac(tan^2theta)^y-1(1+(tan^2theta))^x+yd(tan^2theta)\
&=intlimits_0^frac12pifrac(tan^2theta)^y-1(sec^2theta)^x+y(2tanthetasec^2theta)dtheta\
&=2intlimits_0^frac12pi(sintheta)^2y-2+1(cos theta)^-(2y-2+1)+(2x+2y)-2dtheta\
&= 2 intlimits_0^frac12pi (cos theta)^2x-1(sintheta)^2y-1dtheta\
endalign
Now can you think of the next substitution? (Hint: look at the limits and the exponents)
answered Jul 16 at 4:36
Karn Watcharasupat
3,8172426
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
From the first to second,
beginalign
phi(x,y) &=intlimits_0^inftyfracv^y-1(1+v)^x+ydv\
&=intlimits_0^inftyfrac(tan^2theta)^y-1(1+(tan^2theta))^x+yd(tan^2theta)\
&=intlimits_0^frac12pifrac(tan^2theta)^y-1(sec^2theta)^x+y(2tanthetasec^2theta)dtheta\
&=2intlimits_0^frac12pi(sintheta)^2y-2+1(cos theta)^-(2y-2+1)+(2x+2y)-2dtheta\
&= 2 intlimits_0^frac12pi (cos theta)^2x-1(sintheta)^2y-1dtheta\
endalign
Now can you think of the next substitution? (Hint: look at the limits and the exponents)
answered Jul 16 at 4:36
Karn Watcharasupat
3,8172426
From the first to second,
beginalign
phi(x,y) &=intlimits_0^inftyfracv^y-1(1+v)^x+ydv\
&=intlimits_0^inftyfrac(tan^2theta)^y-1(1+(tan^2theta))^x+yd(tan^2theta)\
&=intlimits_0^frac12pifrac(tan^2theta)^y-1(sec^2theta)^x+y(2tanthetasec^2theta)dtheta\
&=2intlimits_0^frac12pi(sintheta)^2y-2+1(cos theta)^-(2y-2+1)+(2x+2y)-2dtheta\
&= 2 intlimits_0^frac12pi (cos theta)^2x-1(sintheta)^2y-1dtheta\
endalign
Now can you think of the next substitution? (Hint: look at the limits and the exponents)
answered Jul 16 at 4:36
Karn Watcharasupat
3,8172426
edited Jul 16 at 4:43
answered Jul 16 at 4:36
Karn Watcharasupat
3,8172426
answered Jul 16 at 4:36
Karn Watcharasupat
3,8172426
answered Jul 16 at 4:36
Karn Watcharasupat
3,8172426
3,8172426
add a comment |Â
add a comment |Â
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Search for the derivation of the relation between beta and gamma functions.
– StubbornAtom
Jul 16 at 4:32
Second to third is by the substitution $lambda = (cos x)^2$
– Kavi Rama Murthy
Jul 16 at 5:55
See math.stackexchange.com/questions/514025/….
– StubbornAtom
Jul 16 at 6:15