Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What are the usual ways to follow in order to solve the integrals given below? $$\begin{align*} I&=\int_0^1 \ln\Gamma(x)\,dx\\ J&=\int_0^1 x\ln\Gamma(x)\,dx \end{align*}$$

share|improve this question
2  
A quick google search reveals the following paper: google.com/…. These appear to be very serious and tedious integrals to compute. –  nullUser Jul 5 '12 at 20:41
1  
@nullUser: i hope they are not that serious and tedious. Andrew's solution for the first integral is very easy and short. –  Chris's sis Jul 5 '12 at 21:02
    
It looks very easy because he hid all of the difficulty. Computing $\int_0^{\pi}\log \sin x$ is now the serious and tedious integral to compute. –  nullUser Jul 5 '12 at 21:05
1  
@nullUser: i know to compute that integral very easily. Yes, Andrew did the hard part. –  Chris's sis Jul 5 '12 at 21:16
1  
Here is a method of that integral without reflection formula math.stackexchange.com/questions/130621/… –  userNaN Jul 5 '12 at 21:50

5 Answers 5

up vote 4 down vote accepted

As an addendum of sorts to the previous answers, there is the identity

$$\mathrm{logG}(z+1)=\frac{z}{2}\log(2\pi)-\frac{z(z+1)}{2}+z\log\Gamma(z+1)-z(\log\,z-1)-\int_0^z \log\Gamma(t)\,\mathrm dt$$

where $\mathrm{logG}(z)$ is the logarithm of the Barnes function (double gamma function) $G(z)$, the function that satisfies the functional equation $G(z+1)=\Gamma(z)G(z)$. (Barnes proved this identity in his paper, where he introduced the function now named after him.) For $n$ an integer, $G(n)$ can be expressed as

$$G(n)=\prod_{k=1}^{n-2} k!$$

Thus, to evaluate $\int_0^1 \log\Gamma(t)\,\mathrm dt$, we have

$$\begin{align*} \mathrm{logG}(2)&=\frac{1}{2}\log(2\pi)-1+\log\Gamma(2)-(\log\,1-1)-\int_0^1 \log\Gamma(t)\,\mathrm dt\\ 0&=\frac{1}{2}\log(2\pi)-\int_0^1 \log\Gamma(t)\,\mathrm dt \end{align*}$$

and you obtain the same solution as Andrew.


For the integral $\int_0^1 t\log\Gamma(t)\,\mathrm dt$, integration by parts and taking appropriate limits yields the identity

$$\int_0^1 t\log\Gamma(t)\,\mathrm dt=-\frac12\int_0^1 t^2\,\psi(t)\,\mathrm dt$$

Now, Victor Adamchik, in a paper on negative-order polygamma functions (the same sort of functions that appear in Argon's answer), gives the identity

$$\begin{split}&\int_0^z x^n \psi(x) \,\mathrm dx=\\&(-1)^n\left(\frac{B_{n+1} H_n}{n+1}-\zeta^\prime(-n)\right)+\sum_{k=0}^n (-1)^k \binom{n}{k} z^{n-k} \left(\zeta^\prime(-k,z)-\frac{B_{k+1}(z) H_k}{k+1}\right)\end{split}$$

where $B_n$ and $B_n(z)$ are the Bernoulli numbers and polynomials, $H_n=\sum_{j=1}^n\frac1{j}$ is a harmonic number, and $\zeta^\prime(s,a)=\left.\frac{\mathrm d}{\mathrm dt}\zeta(t,a)\right|_{t=s}$ is the derivative of the Hurwitz zeta function.

For $z=1$, the identity simplifies nicely:

$$\int_0^1 x^n \psi(x) \,\mathrm dx=\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}\left(\zeta^\prime(-k)-\frac{B_{k+1} H_k}{k+1}\right)$$

Taking $n=2$, and using the special values $\zeta^\prime(0)=-\frac12\log(2\pi)$ and $\zeta^\prime(-1)=\frac1{12}-\log\,A$, where $A$ is the Glaisher-Kinkelin constant, we finally obtain

$$\int_0^1 x^2 \psi(x) \,\mathrm dx=2\log\,A-\frac12\log(2\pi)$$

and thus

$$\int_0^1 t\log\Gamma(t)\,\mathrm dt=\frac14\log(2\pi)-\log\,A$$

share|improve this answer
    
nice identity! (+1) –  Chris's sis Jul 8 '12 at 6:48
    
Just in case you haven't seen it, I added something for the second integral. –  J. M. Jul 11 '12 at 11:07
    
thanks for your answer! –  Chris's sis Jul 11 '12 at 11:11

As for the first integral, one can use the Euler's reflection formula $\Gamma(1-z) \; \Gamma(z) = {\pi \over \sin{\pi z}}\;$: $$ I=\frac12\int_0^1 ( \log \Gamma(x)+\log \Gamma(1-x))\; dx= \frac12\int_0^1 \log \frac{\pi} {\sin{\pi x}} dx= $$ $$ \frac12\int_0^1 (\log {\pi}-\log {\sin{\pi x}})\; dx= \frac12\log {\pi}-\frac1{2\pi}\int_0^\pi \log {\sin{x}}\; dx= $$ $$ \frac12\log {\pi}-\frac1{2\pi}(-\pi \log 2)=\frac{1}{2} \log 2 \pi. $$ The last integral is well known Gauss integral.

share|improve this answer

As for $J$, another way is to try to use the Fourier series for $\ln\Gamma(x)$ discovered by E.E. Kummer in 1847:

$$\ln\Gamma(x)=\frac{\ln 2\pi}{2}+\sum_{n=1}^{\infty}\frac{\cos 2\pi nx}{2n}+\sum_{n=1}^{\infty}\frac{(\gamma+\ln 2\pi n)\sin 2\pi nx}{n\pi}\,(0<x<1)$$

where $\gamma=0.577\dots$ is Euler's constant

Let's multiply this equality by $x$ and integrate from $0\text{ to }1$.

Integrals on the right side:

$$\begin{align*} &\int_{0}^{1}x\,dx=\frac{1}{2}\\ &\int_{0}^{1}x\cos 2\pi nx\,dx=0\\ &\int_{0}^{1}x\sin 2\pi nx\,dx=-\frac{1}{2\pi n} \end{align*}$$ Thus, $$\begin{align*}\int_{0}^{1}x\ln\Gamma(x)&=\frac{\ln 2\pi}{4}-\frac{\gamma}{2\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}-\frac{1}{2\pi^2}\sum_{n=1}^{\infty}\frac{\ln 2\pi n}{n^2}\\&=\frac{\ln 2\pi}{4}-\frac{\gamma}{12}-\frac{1}{2\pi^2}\sum_{n=1}^{\infty}\frac{\ln 2\pi n}{n^2}\end{align*}$$ if I am not mistaken. I don't know can this be simplified further.

share|improve this answer
1  
it leads to the answer written above by Argon since $\sum_{n=1}^\infty\frac{\log n}{n^2}$ can be expressed through the Glaisher's constant en.wikipedia.org/wiki/Glaisher–Kinkelin_constant –  Andrew Jul 6 '12 at 15:42

The integral $I$ was mentioned on chat recently, and my solution is different than those given before.

Since $x\Gamma(x)=\Gamma(x+1)$, we have $$ \int_0^n\log(\Gamma(x))\,\mathrm{d}x+\int_0^n\log(x)\,\mathrm{d}x =\int_1^{n+1}\log(\Gamma(x))\,\mathrm{d}x\tag{1} $$ Subtracting $\int_1^n\log(\Gamma(x))\,\mathrm{d}x$ from $(1)$ gives $$ \int_0^1\log(\Gamma(x))\,\mathrm{d}x+\int_0^n\log(x)\,\mathrm{d}x =\int_n^{n+1}\log(\Gamma(x))\,\mathrm{d}x\tag{2} $$ Stirling's approximation says $$ \log(\Gamma(x))=x\log(x)-x-\frac12\log(x)+\frac12\log(2\pi)+o(1)\tag{3} $$ Integrating $(3)$ between $n$ and $n+1$ yields $$ \begin{align} &\int_n^{n+1}\log(\Gamma(x))\,\mathrm{d}x\\ &=\left[\frac12x^2\log(x)-\frac14x^2-\frac12x^2-\frac12x\log(x)+\frac12x\right]_n^{n+1}+\frac12\log(2\pi)+o(1)\\ &=n\log(n)-n+\frac12\log(2\pi)+o(1)\tag{4} \end{align} $$ Furthermore, $$ \int_0^n\log(x)\,\mathrm{d}x=n\log(n)-n\tag{5} $$ In light of $(2)$, subtracting $(5)$ from $(4)$ gives $$ \begin{align} \int_0^1\log(\Gamma(x))\,\mathrm{d}x &=\frac12\log(2\pi)+o(1)\\ &=\frac12\log(2\pi)\tag{6} \end{align} $$

share|improve this answer
    
Really nice! (+1). I like to see many solutions to each problem. –  Chris's sis Jul 15 '13 at 7:49

By parts, we have $$J=\int_0^1 x\log \Gamma(x) \, dx=\left[x\psi^{(-2)}(x)\right]_0^1-\int_0^1 \psi^{(-2)}(x)\, dx=\psi^{(-2)}(1)-\psi^{(-3)}(1)=I-\psi^{(-3)}(1)=\frac{1}{4}\log (\frac{2\pi}{A^4})$$

where $A \approx 1.28$ is Glaisher's constant.

share|improve this answer
1  
This is not a solution, for my taste. You just referenced to the values of $\psi$ function which known to the narrow circle of specialists. In order to get I usual solution (which uses only basic facts obout gamma function) one need to repeat some part of Glaisher's work. –  userNaN Jul 5 '12 at 22:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.