# Thread: Gamma Function to evaluate integral - natural logarithm

1. ## Gamma Function to evaluate integral - natural logarithm

Hi,

I've been given the following question in a paper:

Use the Gama function to evaluate the following integral:

Having only briefly covered the Gamma function and only ever coming across questions in the format of

I've done some research on the gamma function and have even had a look over my log rules but I'm really struggling with how to get the integral into the above format.

Any help in where to go with this would be greatly appreciated.

Thank you,

Eloise

2. ## Re: Gamma Function to evaluate integral - natural logarithm

Originally Posted by eloisey
Hi,

I've been given the following question in a paper:

Use the Gama function to evaluate the following integral:

Having only briefly covered the Gamma function and only ever coming across questions in the format of

I've done some research on the gamma function and have even had a look over my log rules but I'm really struggling with how to get the integral into the above format.

Any help in where to go with this would be greatly appreciated.

Thank you,

Eloise
Write this as \displaystyle \begin{align*} \int_0^1{\left(\ln{x}\right)^\frac{3}{2}\,dx} = \int_0^1{\frac{x\left(\ln{x}\right)^{\frac{3}{2}}} {x}\,dx} \end{align*}.

Then make the substitution \displaystyle \begin{align*} u = \ln{x} \implies du = \frac{1}{x}\,dx \end{align*}, note that this means \displaystyle \begin{align*} x = e^u \end{align*} and that when \displaystyle \begin{align*} x = 0, u \to -\infty \end{align*} and when \displaystyle \begin{align*} x = 1, u = 0 \end{align*} and the integral becomes

\displaystyle \begin{align*} \int_0^1{\frac{x\left(\ln{x}\right)^{\frac{3}{2}}} {x}\,dx} &= \int_{-\infty}^0{e^uu^{\frac{3}{2}}\,du} \end{align*}

and make a further substitution \displaystyle \begin{align*} t = -u \implies dt = -du \end{align*}, and note that when \displaystyle \begin{align*} u \to -\infty, t \to \infty \end{align*} and when \displaystyle \begin{align*} u = 0, t = 0 \end{align*} and the integral becomes

\displaystyle \begin{align*} \int_{-\infty}^0{e^uu^{\frac{3}{2}}\,du} &= \int_{\infty}^0{e^{-t}\left(-t^{\frac{3}{2}}\right)\left(-dt\right)} \\ &= \int_{\infty}^{0}{e^{-t}t^{\frac{3}{2}}\,dt} \\ &= -\int_0^{\infty}{e^{-t}t^{\frac{3}{2}}\,dt} \\ &= -\int_0^{\infty}{e^{-t}t^{\frac{5}{2} - 1}\,dt} \end{align*}

which is now in the required form.