Question

How to solve ∫ ln(x!) dx lower bound 0 and upper bond 1?​

Answer 1

We have to evaluate the integral:

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

We know that:

• $x! = \Gamma(x + 1)$
• $\Gamma(x+1) = x \cdot \Gamma(x)$

Therefore, the given integral is equal to:

${ I= \int \limits_0^1 \ln(\Gamma(x+1)) dx}$

${ I= \int \limits_0^1 \ln(x \cdot\Gamma x) dx}$

Now make use of logarithmic property:

• $\ln(x\cdot y) = \ln(x) + \ln(y)$

${ I= \int \limits_0^1 \ln(x) + \ln(\Gamma x) dx}$

${ I= \int \limits_0^1 \ln(x) dx + \int\limits_0^1 \ln(\Gamma x) dx}$

Now, value of $\int\limits_0^1 \ln(x) dx = -1$. You may refer to the solution on this línk in case you are wondering: (brainly.in/question/47449571).

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

(Say it equation 1)

Also we know that,

• $\int \limits_a^bf(x)dx = \int \limits_a^bf(a + b - x)dx$

Therefore,

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

. . . and as a result, equation 1 now becomes:

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

(Say it equation 2)

Adding equation 1 and 2 and dividing by 2, we get:

${ I= - 1 + \dfrac12\int\limits_0^1 \ln(\Gamma( 1 - x)) dx + \dfrac12\int\limits_0^1 \ln(\Gamma x) dx }$

${ I= - 1 + \dfrac12\left[ \int\limits_0^1 \ln(\Gamma( 1 - x)) + \ln(\Gamma x) dx\right] }$

${ I= - 1 + \dfrac12\left[ \int\limits_0^1 \ln(\Gamma x \cdot\Gamma( 1 - x)) dx\right] }$

Now, for 0 < x < 1, we have a formula called Euler's Reflection Formula according to which, $\Gamma x \cdot \Gamma(1-x) = \dfrac{\pi}{\sin x\pi}$.

By using this, our integral changes to:

${ I= - 1 + \dfrac12\left[ \int\limits_0^1 \ln\left[ \dfrac{\pi}{ \sin x\pi} \right] dx\right] }$

${ I= - 1 + \dfrac12\left[ \int\limits_0^1 \ln \pi - \ln(\sin x\pi) dx\right] }$

${ I= - 1 + \dfrac12 \int\limits_0^1 \ln \pi dx - \dfrac12\int\limits_0^1 \ln(\sin x\pi) dx }$

${ I= - 1 + \dfrac{ \ln \pi}2 - \dfrac12\int\limits_0^1 \ln(\sin x\pi) dx }$

For the integral $\int\limits_0^1 \ln(\sin \pi x) dx$, substitute $t = \pi x$. This implies that $dx = \dfrac{dt}{\pi}.$

$\displaystyle{ I= - 1 + \dfrac{ \ln \pi}2 - \dfrac12\int\limits_0^ \pi \dfrac{\ln(\sin t)}{ \pi} dt }$

$\displaystyle{ I= - 1 + \dfrac{ \ln \pi}2 - \dfrac1{2 \pi}\int\limits_0^ \pi \ln(\sin t) dt }$

Now, $\int\limits_0^\pi \ln(\sin t)dt = (-\pi \log 2)$.

(Proof for this is very lengthy, better to learn the formula as it is.)

$\displaystyle{ I= - 1 + \dfrac{ \ln \pi}2 - \dfrac1{2 \pi}( - \pi \ln 2) }$

$\displaystyle{ I= - 1 + \dfrac{ \ln \pi}2 + \dfrac{\ln 2}{2} }$

$\displaystyle{ I= - 1 + \ln \sqrt{\pi} + \ln \sqrt{2} }$

$\displaystyle{ I= - 1 + \ln \sqrt{2\pi} }$

So our final answer is:

$\underline{ \boxed{ \int \limits_0^1 \ln(x!)dx = \ln \sqrt{2\pi} - 1} }$