Gamma Integral

∫−∞∞dxΓ(α+x)Γ(β−x)=2α+β−2Γ(α+β−1);ℜ(α+β)>1 \large \int_{-\infty}^{\infty} \dfrac{ \mathrm{d} x}{\Gamma(\alpha +x) \Gamma (\beta - x)} = \dfrac{2^{\alpha + \beta -2}}{\Gamma(\alpha + \beta -1)} \quad ; \quad \Re (\alpha + \beta) > 1

Prove the equation above without using the method of residues.

Notation: Γ(⋅) \Gamma(\cdot) denotes the gamma function.

This is a part of the set Formidable Series and Integrals.

#Calculus

Note by Ishan Singh
5 years, 3 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Let ξ=∫−∞∞dxΓ(α+x)Γ(β−x)\displaystyle \xi=\int_{-\infty}^{\infty}\frac{dx}{\Gamma(\alpha +x)\Gamma(\beta -x)} Substitute α+x−1=u\displaystyle \alpha+x-1=u and then for convenience let α+β−2=n\displaystyle \alpha+\beta-2=n , hence we get ξ=∫−∞∞duΓ(u+1)Γ(n+1−u)\displaystyle \xi=\int_{-\infty}^{\infty}\frac{du}{\Gamma(u+1)\Gamma(n+1-u)}

Now using that Γ(z+1)=zΓ(z)\displaystyle \Gamma(z+1)=z\Gamma(z) and similarly Γ(n+1−u)=Γ(1−u)∏k=0n(k−u)=(−1)nΓ(1−u)∏k=0n(u−k)\displaystyle \Gamma(n+1-u)=\Gamma(1-u)\prod_{k=0}^n(k-u)=(-1)^n\Gamma(1-u)\prod_{k=0}^n(u-k)

We get ξ=∫−∞∞(−1)nduΓ(u)Γ(1−u)∏k=0n(u−k)\xi=\int_{-\infty}^{\infty}\frac{(-1)^n du}{\displaystyle \Gamma(u)\Gamma(1-u)\prod_{k=0}^n(u-k)}

Now using the Euler's Reflection formula that Γ(u)Γ(1−u)=πcsc⁡(πu)\displaystyle \Gamma(u)\Gamma(1-u)=\pi\csc(\pi u) and the partial fraction decomposition as 1∏k=0n(u−k)=∑k=0naku−k\displaystyle \frac{1}{\displaystyle \prod_{k=0}^n(u-k)}=\sum_{k=0}^n\frac{a_k}{u-k} for some ai∈R∀ i∈{0,1,⋯ ,n}\displaystyle a_i\in\Bbb{R}\forall\space i\in\{0,1,\cdots,n\}. Computing explicitly we get ak=(−1)n−kk!(n−k)!\displaystyle a_k=\frac{(-1)^{n-k}}{k!(n-k)!}. Using this and interchanging the summation and integral signs we get ξ=1πn!∑k=0n(−1)n(−1)n+k(nk)∫−∞∞sin⁡(πu)u−kdu\displaystyle \xi=\frac{1}{\pi n!}\sum_{k=0}^n(-1)^n(-1)^{n+k}\binom{n}{k}\int_{-\infty}^{\infty}\frac{\sin(\pi u)}{u-k}du

Now substituting u−k=t\displaystyle u-k=t in the integral we get =1πn!∑k=0n(−1)k(nk)∫−∞∞sin⁡(πt+πk)tdt\displaystyle =\frac{1}{\pi n!}\sum_{k=0}^n(-1)k\binom{n}{k}\int_{-\infty}^{\infty}\frac{\sin(\pi t+\pi k)}{t}dt

But sin⁡(πt+πk)=(−1)ksin⁡(πt)\displaystyle \sin(\pi t+\pi k)=(-1)^k\sin(\pi t). Using this alongwith the substitution πt=y\pi t=y in the integral we get that ξ=1πn!∑k=0n(nk)∫−∞∞sin⁡yydy\displaystyle \xi=\frac{1}{\pi n!}\sum_{k=0}^n\binom{n}{k}\int_{-\infty}^{\infty}\frac{\sin y}{y}dy

Now using a well know result that ∫−∞∞sin⁡xxdx=π\displaystyle \int_{-\infty}^{\infty}\frac{\sin x}{x}dx=\pi we get ξ=1πn!∑k=0n(nk)π\displaystyle \xi=\frac{1}{\pi n!}\sum_{k=0}^n\binom{n}{k}\pi

Using the Binomial theorem we have that ∑k=0n(nk)=2n\displaystyle \sum_{k=0}^n\binom{n}{k}=2^n and hence ξ=2nn!\displaystyle \xi=\frac{2^n}{n!}

Now recall that n=α+β−2\displaystyle n=\alpha+\beta-2 and that Γ(n+1)=n!\displaystyle \Gamma(n+1)=n!. Hence ξ=∫−∞∞dxΓ(α+x)Γ(β−x)=2α+β−2Γ(α+β−1)\displaystyle \xi=\int_{-\infty}^{\infty}\frac{dx}{\Gamma(\alpha +x)\Gamma(\beta -x)}=\frac{2^{\alpha+\beta-2}}{\Gamma(\alpha+\beta-1)}

Q.E.D\mathbf{\color{#20A900}{\text{Q.E.D}}}

Rohan Shinde - 1 year, 6 months ago

Log in to reply

You have assumed that α+β−2\alpha + \beta - 2 is a non negative integer, which is not necessarily true.

Ishan Singh - 7 months, 3 weeks ago

I wonder if it is possible to prove it without contour, well here is its solution, just put n=0n=0 at the end. @Ishan Singh

Tanishq Varshney - 5 years, 3 months ago

Log in to reply

I haven't tried the general question, but this special case does have an elementary solution.

Ishan Singh - 5 years, 3 months ago
×

Problem Loading...

Note Loading...

Set Loading...