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Math Help - Gamma Function to evaluate integral - natural logarithm

  1. #1
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    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:

    Gamma Function to evaluate integral - natural logarithm-gamma-2.jpg
    Having only briefly covered the Gamma function and only ever coming across questions in the format of

    Gamma Function to evaluate integral - natural logarithm-gamma-1.jpg
    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
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    Re: Gamma Function to evaluate integral - natural logarithm

    Quote Originally Posted by eloisey View Post
    Hi,

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

    Use the Gama function to evaluate the following integral:

    Click image for larger version. 

Name:	gamma 2.JPG 
Views:	6 
Size:	9.6 KB 
ID:	23574
    Having only briefly covered the Gamma function and only ever coming across questions in the format of

    Click image for larger version. 

Name:	gamma 1.JPG 
Views:	1 
Size:	9.5 KB 
ID:	23573
    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.
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