Question

Solve the integral:

$\int \limits_0^ 1 \ln[\Gamma( x)]dx$
​

Answer 1

$\sf I = \int \limits_{0}^{1} ln [ \gamma (x) ] dx$

$\longrightarrow$ Here, We can use identity :$→ \int \limits_{0}^{a} f(x) dx = \int \limits_{0}^{a} f(a - x) dx$

So Use it in given Integral,

$\sf I = \int \limits_{0}^{1} ln [\gamma(1-x)] dx$

$\sf 2I = \int \limits_{0}^{1} ln [\gamma (x)] dx + \int \limits_{0}^{1} ln [\gamma ( 1 - x )] dx$

$\sf 2l = \int \limits_{0}^{1} [\gamma (x) ] + ln [ \gamma (1 - x) ] dx$

$\longrightarrow$ Now, we can here apply $\sf ln \: \: a + ln \: \: b = ln \: \: ab$

$\sf 2I = \int \limits_{0}^{1} In [\gamma (x)] . \gamma (1 - x) dx$

$\longrightarrow$ We know that, $\gamma (x) . \gamma (1 - x) = \frac{π}{sin(πx)}$

$\sf 2I = \int \limits_{0}^{1} In \bigg[\frac{π}{sin(πx)} \bigg] dx$

$\longrightarrow$ $\sf In \bigg( \frac{a}{b} = ln\: \: a - In \: \: b$

$\sf 2I = \int \limits_0^1 In \: \: π \: \: dx - \int \limits_0^1 In [sin(πx)] dx$

:⟹ πx = t

:⟹ πdx = dt

:⟹ x → 0 , t → 0

⟹ x → 1 , t → π

$\sf 2I = In \: \: π . x \bigg| _{0}^{1} - \int \limits_{0}^{π} In (sin \: t ) .\frac{dt}{π}$

$\longrightarrow$ Use : $\sf → \int \limits_0^a f(t) dt = \int \limits_0^a f(x)dx$

Hence,

$\sf 2I = In \: \: π - \frac{1}{π} \int \limits_0^π In (sin \: x ) dx$

$\sf 2I = In \: π - \frac{1}{π} × 2 \int \limits_0^{\frac{π}{2}} In (sin \: x) dx$

Now, focus on, $\sf \int \limits_0^{\frac{π}{2}} In (sin \: x) dx$

$\sf → \frac{-π}{2} In \: 2$

So,

$\sf 2I = In \: π - \frac{2}{π} × \frac{-π}{2} In \: 2$

$\sf 2I = In \: π - \frac{\cancel{2}}{\cancel{π}} × \frac{-\cancel{π}}{\cancel{2}} In \: 2$

$\sf 2I = In \: π + In \: 2$

$\sf 2I = In (2π)$

$\sf I = \frac{1}{2} In (2π)$

$\underline{\boxed{\bf I = In (2π)^{\frac{1}{2}}}}$

We can write it like ↓

$\underline{ \boxed{\bf I = In \sqrt{2π}}}$

I hope it helps you...

Answer 2

Answer:

$\huge \orange{ \frac{1}{2} \log(2\pi) }$

Step-by-step explanation:

To solve this integral first you should know about

$\boxed{\purple {Euler \: reflection \: formula}} \\for \: 0 < x < 1 \\ \\ \Gamma(x) \Gamma(1- x) = \frac{\pi}{ \sin(\pi x) }$

Proof of this formula is very lengthy and tedious so just use it here .

Another property of integral which I will be using frequently is

$\int_{a} ^{b} f(x)dx = \int_{a} ^{b} f(a + b - x)dx \\ \implies \: \\ \\ \int_{0} ^{1} f(x)dx = \int_{0} ^{1} f(1 - x)dx$

Now take log both sides of Euler reflection formula

$log( \Gamma(x) \Gamma(1- x)) = log( \frac{\pi}{ \sin(\pi x) } ) \\ \\ \implies\\log(\Gamma(x)) + log(\Gamma(1- x) )= log( {\pi}) - log( { \sin(\pi x) } )$

Integrate both sides 0 to 1 ,

$\red{LHS} \\ \\ \int_{0} ^{1} log(\Gamma(x))dx + \int_{0} ^{1} log(\Gamma(1- x) )dx \\ \\ integral \: property \\ = \int_{0} ^{1} log(\Gamma(x))dx + \int_{0} ^{1} log(\Gamma(1- (1 - x)) )dx \\ \\ = \int_{0} ^{1} log(\Gamma(x))dx + \int_{0} ^{1} log(\Gamma(x) )dx \\ \\ = 2 \int_{0} ^{1} log(\Gamma(x))dx$

Now

$\red{RHS}$

$\int _{0} ^{1} log( {\pi}) dx- \int _{0} ^{1} log( { \sin(\pi x) } )dx \\ \: put \: \: \pi x = t \\ = log( {\pi}) - \frac{1}{\pi} \int _{0} ^{\pi} log( { \sin( t) } )dt$

Let

$I = \int _{0} ^{\pi} log( { \sin( t) } )dt \\ \\ = \int _{0} ^{\pi \over2} log( { \sin( t) } )dt + \int _{\pi \over2} ^{\pi} log( { \sin( t) } )dt \\ \\ replace \: \: t \: \: by \: \: t - \frac{\pi}{2} \: in \: second \: term \\ \\ = \int _{0} ^{\pi \over2} log( { \sin( t) } )dt + \int _{0} ^{\pi \over2} log( { \cos( t) } )dt \\ \\ = \int _{0} ^{\pi \over2} log( { \sin(t) \cos( t) } )dt \\ \\ = \int _{0} ^{\pi \over2} log( { 2\sin(t) \cos( t) \over2 } )dt \\ \\ = \int _{0} ^{\pi \over2} log( { \sin(2t) } )dt - \int _{0} ^{\pi \over2} log(2) dt \\ \\ replace \: \: 2t \: \: by \: \: t \: in \: first \: term\\ \\ = \frac{1}{2} \int _{0} ^{\pi} log( { \sin(t) } )dt - \frac{\pi}{2} log(2) \\ \\ \implies \\ \\ I = \frac{I}{2} - \frac{\pi}{2} log(2) \\ \\ \implies \: I = - \pi \: log(2)$

So RHS now become

$= log(\pi) - \frac{1}{\pi} I \\ \\ = log(\pi) - \frac{1}{\pi} ( - \pi \: log(2) ) \\ \\ = log(\pi) - log(2) \\ \\ = log(2\pi)$

Now finally

$LHS = RHS \\ \\ \implies \: \\ \\ 2 \int_{0} ^{1} log(\Gamma(x))dx = log(2\pi) \\ \\ \implies \: \int_{0} ^{1} log(\Gamma(x))dx = \frac{1}{2} log(2\pi)$