An Interesting Integral

**sbhatnagar** · January 13th 2013, 11:05 PM

Prove that

$\int_0^1 \ln \left( \Gamma(1+x)\right)dx = \frac{1}{2}\ln(2\pi)-1$

**sbhatnagar** · January 16th 2013, 09:53 PM

This Problem is not as difficult as it looks. Use the property $\Gamma(1+x)=x\Gamma(x)$ :

$\begin{aligned} \int_0^1 \ln(\Gamma(1+x))dx &= \int_0^1 \ln(x\Gamma(x))dx \\ &= \int_0^1 \ln x dx+\int_0^1\ln(\Gamma(x))dx\end{aligned}$

The first integral equals -1.

$\begin{aligned} \int_0^1 \ln(\Gamma(1+x))dx &= \int_0^1\ln(\Gamma(x))dx-1\end{aligned}$

Let $I=\int_0^1 \ln(\Gamma(x)) dx$ -------(1)

A property of definite integral says that

$\int_0^a f(x)dx=\int_0^a f(a-x) dx$

therefore

$I=\int_0^1 \ln(\Gamma(1-x))dx$ --------(2)

Add (1) and (2):

$2I = \int_0^1 \ln(\Gamma(x) \Gamma(1-x))dx$

$= \int _0^1 \ln \left( \frac{\pi}{\sin(\pi x)}\right)dx$ (from Euler's Reflection Formula!

)

$=\ln(\pi)\int_0^1 dx - \int_0^1 \ln(\sin (\pi x))dx$

Recall that $\int_0^1 \ln(\sin(\pi x)) dx= -\ln(2)$ .

Therefore $2I= \ln(2\pi)$

$I=\frac{1}{2}\ln(2\pi)$

$\int_0^1 \ln(\Gamma(1+x))dx=\frac{1}{2}\ln(2\pi)-1$

**zaidalyafey** · January 18th 2013, 02:40 AM

How would you prove the convergence of this integral ... ?

**SworD** · January 18th 2013, 12:33 PM

Originally Posted by zaidalyafey

How would you prove the convergence of this integral ... ?

The integrand is real and finite in that interval.

**SworD** · January 19th 2013, 04:20 PM

Here is another way to evaluate the integral. Write it out as a Riemann sum, the old-school way:

$\int_{a}^{a+1} f(x) dx = \lim_{n \to \infty} \frac{1}{n} \left (f(a) + f(a + \frac{1}{n}) + f(a + \frac{2}{n}) + f(a + \frac{3}{n}) .... + f(a + \frac{n-1}{n}) \right )$

$\int_0^1 \ln \left( \Gamma(1+x)\right)dx = \int_1^2 \ln \left( \Gamma(x)\right)dx$

So

$\int_1^2 \ln(\Gamma(x))dx = \lim_{n \to \infty} \frac{1}{n} \left (\ln(\Gamma(1)) + \ln(\Gamma(1 + \frac{1}{n})) + \ln(\Gamma(1 + \frac{2}{n})) + \ln(\Gamma(1 + \frac{3}{n})) .... + \ln(\Gamma(1 + \frac{n-1}{n})) \right )$

$= \lim_{n \to \infty} \frac{1}{n} \ln\left ( \Gamma(1)\Gamma(1 + \frac{1}{n})\Gamma(1 + \frac{2}{n})\Gamma(1 + \frac{3}{n}) ....\Gamma(1 + \frac{n-1}{n}) \right )$

This exactly fits the format for the multiplication theorem for the gamma function! So

$I = \int_1^2 \ln \left( \Gamma(x)\right)dx = \lim_{n \to \infty} \frac{1}{n}\ln \left ((2\pi)^\frac{n-1}{2}\cdot n^{\frac{1}{2}-n}\cdot \Gamma(n))\right )$

$e^I = \lim_{n \to \infty} \left ((2\pi)^\frac{n-1}{2}\cdot n^{\frac{1}{2}-n}\cdot \Gamma(n))\right )^\frac{1}{n}$

$= \lim_{n \to \infty} \left ((2\pi)^\frac{n-1}{2n}\cdot n^{\frac{1}{2n}-1}\cdot ((n-1)!)^\frac{1}{n}\right )$

$= \sqrt{2\pi }\lim_{n \to \infty} \left (\frac{n^{\frac{1}{2n}}}{n}\cdot ((n-1)!)^\frac{1}{n}\right )$

$= \sqrt{2\pi } \cdot \lim_{n \to \infty} \left ( n^{\frac{1}{2n}} \right ) \cdot \lim_{n \to \infty} \left ( \frac{((n-1)!)^{\frac{1}{n}}}{n} \right )$

$= \frac{\sqrt{2\pi}}e$

$I = \ln{\frac{\sqrt{2\pi}}e}$
$= \frac{1}{2}\ln(2\pi) - 1$

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