Is it possible to simplify Γ(110)Γ(215) Γ(715)\frac{\Gamma\left(\frac{1}{10}\right)}{\Gamma\left(\frac{2}{15}\right)\ \Gamma\left(\frac{7}{15}\right)}?

Is it possible to simplify this expression?
Γ(110)Γ(215) Γ(715)
Is there a systematic way to check ratios of Gamma-functions like this for simplification possibility?

Answer

Amazingly, this can be greatly simplified. I’ll state the result first:

Γ(110)Γ(215)Γ(715)=5+131/1026/5π

The result follows first from a version of Gauss’s multiplication formula:

Γ(3z)=12π33z1/2Γ(z)Γ(z+13)Γ(z+23)

or, with z=2/15:

Γ(215)Γ(715)=2π31/10Γ(25)Γ(45)

Now use the duplication formula

Γ(2z)=1π22z1Γ(z)Γ(z+12)

or, with z=2/5:

Γ(25)Γ(45)=π21/5Γ(910)

Putting this all together, we get

Γ(110)Γ(215)Γ(715)=Γ(110)Γ(910)π326/531/10

And now, we may use the reflection formula:

Γ(z)Γ(1z)=πsinπz

With z=1/10, and noting that

sin(π10)=514=15+1

the stated result follows. This has been verified numerically in Wolfram|Alpha.

Attribution
Source : Link , Question Author : X.C. , Answer Author : Ron Gordon

Leave a Comment