Integral Proofs for Consideration: Gamma

I saw these definite integrals in a standard mathematical table, can someone provide proofs to these so that we can all learn the interesting methods to solving these integrals?

  1. \[\Large{\int_{0}^{1}{ (\log { \dfrac{1}{x} })^n }\, dx = n! \quad\quad n > -1 } \]

  2. 01x(11+xkex)dx=γk>0\Large{\int_{0}^{\infty}{\dfrac{1}{x}(\dfrac{1}{1+x^k}-e^{-x})}\, dx = \gamma \quad\quad k>0 }

  3. 0π2cosxdx=(2π)32(Γ(14))2\Large{\int_{0}^{\frac{\pi}{2}}{\sqrt{\cos{x} } } \, dx = \dfrac{(2\pi)^\frac{3}{2} }{ (\Gamma(\frac{1}{4} ) )^2 } }

  4. 01lnΓ(q)dq=ln2π\Large{\int_{0}^{1}{ \ln { \Gamma(q)} }\, dq =\ln{ \sqrt{2\pi}}}

  5. 0W(x)xxdx=22π\Large{\int_{0}^{\infty}{\dfrac{ \operatorname{W}(x)}{x\sqrt{x}} }\, dx =2\sqrt{2\pi}}

  6. einxΓ(α+x)Γ(βx)dx=(2cosn2)α+β2Γ(α+β1)ein2(βα)n<πand(α+β)>1\large{\int_{-\infty}^{\infty}\frac{e^{i nx}}{\Gamma(\alpha+x) \Gamma(\beta -x)}dx = \frac{ \left(2\cos \frac{n}{2} \right)^{\alpha +\beta-2}}{\Gamma(\alpha+\beta-1)}e^{\frac{in}{2}(\beta - \alpha)} \quad |n|<\pi \quad \text{and} \quad \Re(\alpha+\beta)>1}

For clarity:

Γ(x)\large{ \Gamma(x)} =The Gamma Function

lnΓ(x)\large{ \ln { \Gamma(x)}}=The Log-Gamma Function

γ\large{\gamma} =The Euler-Mascheroni Constant

W(x)\large{\operatorname{W}(x)} = Lambert W-Function

ex\large{e^x} = Natural Exponential Function

π\large{\pi} = Pi

Thank you to everyone who helps!

Image Credits: Google

Note by Hussein Hijazi
5 years, 4 months ago

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@Calvin Lin How can I center the links with the functions and constants? It always starts a new paragraph and I believe thats because I closed the LaTeX boundaries, but I did that so I could add a link :S

Hussein Hijazi - 5 years, 4 months ago

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We currently do not allow for linking within Latex.

I've edited the first line accordingly, where it is not centered (and I don't see a strong reason for doing so).

Calvin Lin Staff - 5 years, 4 months ago

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Thanks! Got it organized now :)

Hussein Hijazi - 5 years, 4 months ago

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Hussein Hijazi - 5 years, 4 months ago

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  1. u=ln(x1)x=eudx=euduu=\ln(x^{-1})\to x=e^{-u}\to dx=-e^{-u} du. the boundaries: 010\mid_0^1\to \mid_\infty^0. we get the integral 0un(eu)du=0uneudu=Γ(n+1)=n!\int_\infty^0 u^n(-e^{-u})du=\int_0^\infty u^ne^{-u} du =\Gamma(n+1)=n!

Aareyan Manzoor - 4 years, 8 months ago

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