Proving Identities for 0<x<1: Integrals and Substitution Techniques

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In summary, their hint will show you those two identities are equivalent. You can derive these with a contour integration, but if you have some useful earlier results, that would be handy. What exactly do you know at this point about Beta integrals, gamma functions, and how they relate to sin?never mind those question, i found in my text that Gamma(x)Gamma(1-x)=pi/sin(pi*x).
  • #1
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i need to prove the next identities for 0<x<1:

∫t^(x-1)/(1+t)dt=π/sin(πx)
0

1
∫t^(x-1)(1-t)^(-x)dt=π/sin(πx)
0

for the second one, my text gives me a hint to substitute t=u/(u+1), but i didnt succeed in getting the rhs.
i tried the defintion of B(x,1-x)=Gamma(x)Gamma(1-x)
but i don't know how to proceed from there.

thanks in advance.
 
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  • #2
loop quantum gravity said:
i need to prove the next identities for 0<x<1:

∫t^(x-1)/(1+t)dt=π/sin(πx)
0

1
∫t^(x-1)(1-t)^(-x)dt=π/sin(πx)
0

for the second one, my text gives me a hint to substitute t=u/(u+1), but i didnt succeed in getting the rhs.
i tried the defintion of B(x,1-x)=Gamma(x)Gamma(1-x)
but i don't know how to proceed from there.

thanks in advance.

Their hint will show you those two identities are equivalent.

You can derive these with a contour integration, but if you have some useful earlier results, that would be handy. What exactly do you know at this point about Beta integrals, gamma functions, and how they relate to sin?
 
  • #3
never mind those question, i found in my text that Gamma(x)Gamma(1-x)=pi/sin(pi*x).

i have another couple questions:
prove the identities:
1)Gamma(x/3)Gamma((x+1)/3)Gamma((x+2)/3=(2pi/3^(x-0.5)Gamma(x)
2)[tex]\int_{0}^{\frac{\pi}{2}}\sqrt cos(x) dx=(2\pi)^{3/2}/(\Gamma(1/4))^2[/tex]

about the first, here what i did:
f(x)=(3^x)gamma(x/3)gamma((x+1)/3)gamma((x+2)/3)
f(x+1)=xf(x)
f(x) is log convex then f(x)=f(1)gamma(x)
f(1)=3*gamma(1/3)gamma(2/3)gamma(1)
gamma(1/3)gamma(2/3)=B(1/3,2/3)
my problem is to compute the integral of B(1/3,2/3) where B is the beta function.

about the second question:
[tex]\int_{0}^{\frac{\pi}{2}}\sqrt cos(x) dx=B(1/2,3/4)=[\gamma(1/2)\gamma(3/4)]/\gamma(5/4)[/tex]
i know that gamma(5/4)=1/4gamma(1/4) and gamma(1/2)=sqrt(pi)
but i don't know how to compute: gamma(3/4).

thanks in advance.
 
Last edited:
  • #4
loop quantum gravity said:
i found in my text that Gamma(x)Gamma(1-x)=pi/sin(pi*x).

This identity can be used to solve both your problems.
 
  • #5
ok, another question about gamma function, i have these two integrals:
[tex]I=\int_{0}^{\infty}\frac{dx}{\sqrt(1+x^{\alpha})}[/tex]
for alpha>2
and [tex]J=\int_{0}^{1}\frac{dx}{\sqrt(1-x^{\alpha}}[/tex]
to represent them as B function and to deduce that J=Icos(1/2).

for the first one by substitution t=x^a/(x^a+1) i got that I=(1/a)B(1/a,0.5-1/a), the second integral i tried by the next two represntations and both got me nowhere:
t=x^a/(1-x^a) and t=(x^a+1)/x^a
 

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