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Math Help - Integration - Engineer in need!

  1. #1
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    Integration - Engineer in need!

    Hello people; here's how it goes, i'm writing my masters dissertation at the moment about the mechanical properties of metallic glasses (interesting stuff eh?!!) and i've got to integrate a function. Now, i got 34% on my undergrad calc test so, try as i might, i just can't do it! Please can someone throw me a lifeline!!

    Here's the function to integrate y = (x^4.e^x)/((e^x-1)^2) dx

    Thanks!!
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  2. #2
    Moo
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    Hello,

    You can do an integration by parts :

    u(x)=x^4

    v'(x)=\frac{e^x}{(e^x-1)^2}, which is something like \frac{g'(x)}{g^2(x)}

    You should know an antiderivative for it


    And then, iterate the integration by parts 3 times (I think)
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by Moo View Post
    And then, iterate the integration by parts 3 times (I think)
    That's what I thought too but when you get to \int \frac{4x^3}{e^x-1}\mathrm{d}x I'm not sure you can go on. :/
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    That's what I thought too but when you get to \int \frac{4x^3}{e^x-1}\mathrm{d}x I'm not sure you can go on. :/
    Yeah you cannot I am pretty sure
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  5. #5
    Moo
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    That's what I thought too but when you get to \int \frac{4x^3}{e^x-1}\mathrm{d}x I'm not sure you can go on. :/


    Too bad ^^
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  6. #6
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    It can't be solved, but by takin' \int_0^\infty\frac{x^3}{e^x-1}\,dx one can solve it, by using \zeta(4) result.
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  7. #7
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    Quote Originally Posted by john5115 View Post
    Hello people; here's how it goes, i'm writing my masters dissertation at the moment about the mechanical properties of metallic glasses (interesting stuff eh?!!) and i've got to integrate a function. Now, i got 34% on my undergrad calc test so, try as i might, i just can't do it! Please can someone throw me a lifeline!!

    Here's the function to integrate y = (x^4.e^x)/((e^x-1)^2) dx

    Thanks!!
    You are not an undergraduate now, this is not for a calculus course. So you do what a professional would do, use technology or look it up.

    Quick maths gives the attachment ( Li_n is the polylogarithm)

    RonL
    Attached Thumbnails Attached Thumbnails Integration - Engineer in need!-gash.jpg  
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  8. #8
    Member Danshader's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    You are not an undergraduate now, this is not for a calculus course. So you do what a professional would do, use technology or look it up.

    Quick maths gives the attachment ( Li_n is the polylogarithm)

    RonL
    i am sorry if i am being impolite but could you show how the calculations would go if i don't using technology to solve it.
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  9. #9
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    Ok, so what do you do with these polylogarithm things?

    Thanks
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  10. #10
    Super Member PaulRS's Avatar
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    But you can find a beautiful formula for the improper Integral.

    Note that: <br />
\int_0^\infty  {\frac{{x^{s - 1}  \cdot e^x }}<br />
{{\left( {e^x  - 1} \right)^2 }}dx}  = \int_0^\infty  {\frac{{\tfrac{{x^{s - 1} }}<br />
{{e^x }}}}<br />
{{\left( {1 - \tfrac{1}<br />
{{e^x }}} \right)^2 }}dx}  = \int_0^\infty  {x^{s - 1} \sum\limits_{n = 1}^\infty  {n \cdot e^{ - xn} }  \cdot dx} <br />

    <br />
\int_0^\infty  {\frac{{x^{s - 1}  \cdot e^x }}<br />
{{\left( {e^x  - 1} \right)^2 }}dx}  = \sum\limits_{n = 1}^\infty  {n \cdot \int_0^\infty  {x^{s - 1}  \cdot e^{ - xn}  \cdot dx} } <br />

    Now: <br />
\int_0^\infty  {x^{s - 1}  \cdot e^{ - xn}  \cdot dx}  = \frac{{\Gamma \left( s \right)}}<br />
{{n^s }}<br />

    Thus: <br />
\int_0^\infty  {\frac{{x^{s - 1}  \cdot e^x }}<br />
{{\left( {e^x  - 1} \right)^2 }}dx}  = \Gamma \left( s \right) \cdot \left( {\sum\limits_{n = 1}^\infty  {\tfrac{1}<br />
{{n^{s - 1} }}} } \right) = \Gamma \left( s \right) \cdot \zeta \left( {s - 1} \right)<br />


    In particular: <br />
\int_0^\infty  {\frac{{x^3  \cdot e^x }}<br />
{{\left( {e^x  - 1} \right)^2 }}dx}  = \Gamma \left( 4 \right) \cdot \zeta \left( 3 \right)

    And: <br />
\int_0^\infty  {\frac{{x^4  \cdot e^x }}<br />
{{\left( {e^x  - 1} \right)^2 }}dx}  =\Gamma \left( 5 \right) \cdot \zeta \left( 4 \right)=\frac{4\cdot{\pi^4}}{15}
    Last edited by PaulRS; April 27th 2008 at 11:29 AM.
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  11. #11
    Grand Panjandrum
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    Quote Originally Posted by PaulRS View Post
    But you can find a beautiful formula for the improper Integral.

    Note that: <br />
\int_0^\infty {\frac{{x^{s - 1} \cdot e^x }}<br />
{{\left( {e^x - 1} \right)^2 }}dx} = \int_0^\infty {\frac{{\tfrac{{x^{s - 1} }}<br />
{{e^x }}}}<br />
{{\left( {1 - \tfrac{1}<br />
{{e^x }}} \right)^2 }}dx} = \int_0^\infty {x^{s - 1} \sum\limits_{n = 1}^\infty {n \cdot e^{ - xn} } \cdot dx} <br />

    <br />
\int_0^\infty {\frac{{x^{s - 1} \cdot e^x }}<br />
{{\left( {e^x - 1} \right)^2 }}dx} = \sum\limits_{n = 1}^\infty {n \cdot \int_0^\infty {x^{s - 1} \cdot e^{ - xn} \cdot dx} } <br />

    Now: <br />
\int_0^\infty {x^{s - 1} \cdot e^{ - xn} \cdot dx} = \frac{{\Gamma \left( s \right)}}<br />
{{n^s }}<br />

    Thus: <br />
\int_0^\infty {\frac{{x^{s - 1} \cdot e^x }}<br />
{{\left( {e^x - 1} \right)^2 }}dx} = \Gamma \left( s \right) \cdot \left( {\sum\limits_{n = 1}^\infty {\tfrac{1}<br />
{{n^{s - 1} }}} } \right) = \Gamma \left( s \right) \cdot \zeta \left( {s - 1} \right)<br />


    In particular: <br />
\int_0^\infty {\frac{{x^3 \cdot e^x }}<br />
{{\left( {e^x - 1} \right)^2 }}dx} = \Gamma \left( 4 \right) \cdot \zeta \left( 3 \right)

    And: <br />
\int_0^\infty {\frac{{x^4 \cdot e^x }}<br />
{{\left( {e^x - 1} \right)^2 }}dx} =\Gamma \left( 5 \right) \cdot \zeta \left( 4 \right)=\frac{4\cdot{\pi^4}}{15}
    But we are not asked for definite integrals. If we are doing engineering we don't worry about this hard work we just compute a numerical integral.

    RonL
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  12. #12
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    Quote Originally Posted by Danshader View Post
    i am sorry if i am being impolite but could you show how the calculations would go if i don't using technology to solve it.
    Look up polylogarithm on wikipedia or mathworld, and you will see how they might occur when this thing is integrated by parts.

    RonL
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  13. #13
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    Quote Originally Posted by john5115 View Post
    Ok, so what do you do with these polylogarithm things?

    Thanks
    The same thing that you would do with anyother special function, look them up, use asymtotic forms to investigate behaviour for large arguments, use series to approximate for small arguments, ...

    RonL
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  14. #14
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    well, ol' Captain Black, i think you are over estimating my mathematical ability!

    Still, i'll give it ago!
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