How to solve $\int\limits_0^1 \ln(x!) dx$?

Question

So in the given question, we have to evaluate the given integral which is $\int\limits_0^1 \ln(x!) dx$

My initial thought is to simplify factorial x inside logarithmic function.

$\implies\int\limits_0^1 \ln((x) \cdot(x-1)\cdot(x-2) \cdot\cdot\cdot 3\cdot 2\cdot 1) dx$

$\implies\int\limits_0^1 \ln(x) + \ln(x-1) +\ln(x-2) +...+\ln(2) + \ln(1)dx$

$\implies\int\limits_0^1 \ln(x) dx+ \int\limits_0^1\ln(x-1) dx+\int\limits_0^1\ln(x-2) dx+...+\int\limits_0^1\ln(2)dx + \int\limits_0^1\ln(1)dx$

Does this make any sense?

The another method which I thought:

$\int\limits_0^1 \ln(x!) dx = \int\limits_0^1 \ln(\Gamma(x+1)) dx$

or

$\int\limits_0^1 \ln(x!) dx = \int\limits_0^1 \ln(\int\limits_0^\infty t^x\cdot e^{-t} dt) dx$

Can I proceed in any of the two methods which I used?
Is there any easier/different method to solve it?

For $x\in[0,1]$ the expansion $x!=x(x-1)(x-2)\cdots2\cdot1$ makes no sense; nor does $\ln(x-1)$, etc., since the natural logarithm is not (naturally) defined on the negative reals, which $x-1$ etc. will be. — Barry Cipra, Oct 13 '21 at 17:44
No mathematician nowadays writes $x!$ for non-integer $x$. Maybe in the 19th century they did. Now you should write it in terms of the $\Gamma$ function. — GEdgar, Oct 13 '21 at 17:55

VIVID · Accepted Answer · 2021-10-13T17:53:54.437

1

Hint 0. $$\Gamma(x+1) = x\Gamma(x)$$

Hint 1. $$I = \int_0^1 \ln\Gamma(x+1)dx = \underbrace{\int_0^1\ln(x)dx}_{I_1} + \underbrace{\int_0^1\ln\Gamma(x) dx}_{I_2}$$

Hint 2. $$I_2 = \int_0^1\ln\Gamma(x) dx = \int_0^1\ln\Gamma(1- x) dx \\ 2I_2 = I_2+I_2 = \int_0^1\ln\Gamma(x) dx + \int_0^1\ln\Gamma(1- x) dx = \int_0^1\ln\Gamma(x)\Gamma(1-x) dx$$

Hint 3. $$\Gamma(x)\Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$$

Hint 4. Putting all together, you will have to know how to solve a classical integral $\int_0^\pi \ln\sin x dx$

edited Oct 13 '21 at 17:53

answered Oct 13 '21 at 17:40

VIVID

11,604

May I know how did you get 2I in step 2? – Oct 13 '21 at 17:48
Please, see the edit. @I_am_Math_Student – VIVID Oct 13 '21 at 17:50
How did you get $\int_0^1 \ln(\Gamma(x)) dx = \int_0^1 \ln(\Gamma(1-x)) dx$? – Oct 13 '21 at 18:03
@I_am_Math_Student $\int_a^b f(x)dx = \int_a^b f(a+b - x)dx$, proof is quite easy by substituion. – VIVID Oct 13 '21 at 18:07
Those are some rather big hints. ;-) – Mark Viola Oct 13 '21 at 18:13
1

@MarkViola Yeah, $\sum_i \text{Hint }i = \text{Solution}$ but maybe the reader gets enough from some of them and will proceed by themselves :-) – VIVID Oct 13 '21 at 18:22
A hint could be a comment, but a “rather big hint” should actually answer the question as an answer. – Тyma Gaidash Oct 13 '21 at 18:25
Can anyone explain how is $\frac{1}{2}\int_0^1\log(\sin x\pi)dx = \frac{1}{2π} \int_0^π\log(\sin x) dx$? I got stucked on this step with reference an answer on the question whose duplicate my question has marked been. – Oct 13 '21 at 18:35
@VIVID can you explain me please? – Oct 13 '21 at 18:42
@I_am_Math_Student take $t = x\pi$ then $dx = \frac{dt}{\pi}$ and $t \in [0,\pi]$ when $x\in [0,1]$. – VIVID Oct 13 '21 at 19:02
Last thing, how is $\frac12 \ln(π) + \frac12 \ln(2) = \ln(\sqrt{2π})$? – Oct 13 '21 at 19:21
@VIVID ........ – Oct 13 '21 at 19:22
@I_am_Math_Student $a\ln b = \ln b^a$ and $\ln a + \ln b = \ln ab$. – VIVID Oct 14 '21 at 11:54

How to solve $\int\limits_0^1 \ln(x!) dx$?

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