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# 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. $\Large{\int_{0}^{\infty}{\dfrac{1}{x}(\dfrac{1}{1+x^k}-e^{-x})}\, dx = \gamma \quad\quad k>0 }$

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

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

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

6. $\large{\int_{-\infty}^{\infty}\frac{e^{i nx}}{\Gamma(\alpha+x) \Gamma(\beta -x)}dx = \frac{ \left(2\cos \frac{n}{2} \right)^{\alpha +\beta-2}}{\Gamma(\alpha+\beta-1)}e^{\frac{in}{2}(\beta - \alpha)} \quad |n|<\pi \quad \text{and} \quad \Re(\alpha+\beta)>1}$

For clarity:

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

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

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

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

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

$$\large{\pi}$$ = Pi

Thank you to everyone who helps!

Note by Hussein Hijazi
2 years, 2 months ago

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1. $$u=\ln(x^{-1})\to x=e^{-u}\to dx=-e^{-u} du$$. the boundaries: $$\mid_0^1\to \mid_\infty^0$$. we get the integral $\int_\infty^0 u^n(-e^{-u})du=\int_0^\infty u^ne^{-u} du =\Gamma(n+1)=n!$
· 1 year, 6 months ago

bump · 2 years, 2 months ago

@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 · 2 years, 2 months ago

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). Staff · 2 years, 2 months ago

Thanks! Got it organized now :) · 2 years, 2 months ago

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