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Math Help - Partial fractions and a funky integral

  1. #1
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    Partial fractions and a funky integral

    QUESTION #1:

    Suppose we have a partial fraction as follows:

    \frac{Ax^2+Bx+C}{(Dx+E)(Fx+G)(Hx+I)}

    Can we break that up into partial fractions like so:

    \frac{J}{Dx+E}+\frac{K}{Fx+G}+\frac{L}{Hx+I}

    Where A,B,C,D,E,F,G,H,I,J,K,L are all constants? Or does the whole numerator of the original equation have to be a constant?

    QUESTION #2:

    How in the world would I go about integrating the following function?

    \int\frac{x^5}{e^x-1}dx

    Thanks!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by hatsoff View Post
    QUESTION #1:

    Suppose we have a partial fraction as follows:

    \frac{Ax^2+Bx+C}{(Dx+E)(Fx+G)(Hx+I)}

    Can we break that up into partial fractions like so:

    \frac{J}{Dx+E}+\frac{K}{Fx+G}+\frac{L}{Hx+I}

    Where A,B,C,D,E,F,G,H,I,J,K,L are all constants? Or does the whole numerator of the original equation have to be a constant?
    yes

    QUESTION #2:

    How in the world would I go about integrating the following function?

    \int\frac{x^5}{e^x-1}dx

    Thanks!
    i doubt this integral has a closed-form solution. you can wait to see if Krizalid comes up with anything. or maybe Mathstud can show you how to use series to solve it
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  3. #3
    Moo
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    Hello,

    Quote Originally Posted by hatsoff View Post
    QUESTION #1:

    Suppose we have a partial fraction as follows:

    \frac{Ax^2+Bx+C}{(Dx+E)(Fx+G)(Hx+I)}

    Can we break that up into partial fractions like so:

    \frac{J}{Dx+E}+\frac{K}{Fx+G}+\frac{L}{Hx+I}

    Where A,B,C,D,E,F,G,H,I,J,K,L are all constants? Or does the whole numerator of the original equation have to be a constant?
    Yes you can
    As soon as the degree of the numerator doesn't exceed the denominator's one ^^
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  4. #4
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    The integral makes sense only if x\in[0,\infty).

    Actually, the integral can be tackled by using series.
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    yes
    Neato.

    i doubt this integral has a closed-form solution. you can wait to see if Krizalid comes up with anything. or maybe Mathstud can show you how to use series to solve it
    It's actually the indefinite form of a definite integral taken from one of Mathstud's problems--but I didn't want to spam his thread by asking high-school-level questions.

    What do you mean "closed-form solution"?
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  6. #6
    Moo
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    What about this :

    I=\int \frac{x^5}{e^x-1} dx

    x \rightarrow -x

    I=\int \frac{x^5}{e^{-x}-1} dx

    Multiply above and below by -e^x :

    I=\int \frac{-x^5 e^x}{e^x-1} dx

    -------------

    2I=\int \frac{x^5}{e^x-1} dx+\int \frac{-x^5 e^x}{e^x-1} dx=\int \frac{x^5 (1-e^x)}{e^x-1}=-\int x^5 dx=-\frac{x^6}{6}

    Therefore, I=-\frac{x^6}{12}


    Is there any mistake ? o.O
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Actually, the integral can be tackled by using series.
    yes, i saw that afterwards...
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  8. #8
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    Quote Originally Posted by Moo View Post

    Is there any mistake ? o.O
    Quote Originally Posted by Moo View Post

    What about this :

    I=\int \frac{x^5}{e^x-1} dx

    x \rightarrow -x
    You can't do that, its application only works for definite integrals.
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  9. #9
    Moo
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    I see ! This is definitely a problem
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Moo View Post
    I see ! This is definitely a problem
    It was a valiant effort though
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  11. #11
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    Quote Originally Posted by hatsoff View Post
    QUESTION #1:

    Suppose we have a partial fraction as follows:

    \frac{Ax^2+Bx+C}{(Dx+E)(Fx+G)(Hx+I)}

    Can we break that up into partial fractions like so:

    \frac{J}{Dx+E}+\frac{K}{Fx+G}+\frac{L}{Hx+I}

    Where A,B,C,D,E,F,G,H,I,J,K,L are all constants? Or does the whole numerator of the original equation have to be a constant?

    QUESTION #2:

    How in the world would I go about integrating the following function?

    \int\frac{x^5}{e^x-1}dx

    Thanks!
    Two things

    I have never seen this indefinite integral before but..

    \int_0^{\infty}\frac{u^{x-1}}{e^x-1}dx=\Gamma(x)\zeta(x)=(x-1)!\sum_{n=1}^{\infty}\frac{1}{n^x}

    alternatively

    \int\frac{x^5}{e^x-1}dx=\int\frac{x^4}{1+\frac{x}{2!}+\frac{x^2}{3!}+  ...\frac{x^{n-1}}{n!}}dx

    Choosing the accuracy at which you wish you can tuncate the bottom to a doable amount of terms
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Two things

    I have never seen this indefinite integral before but..

    \int_0^{\infty}\frac{u^{x-1}}{e^x-1}dx=\Gamma(x)\zeta(x)=(x-1)!\sum_{n=1}^{\infty}\frac{1}{n^x}

    alternatively

    \int\frac{x^5}{e^x-1}dx=\int\frac{x^4}{1+\frac{x}{2!}+\frac{x^2}{3!}+  ...\frac{x^{n-1}}{n!}}dx

    Choosing the accuracy at which you wish you can tuncate the bottom to a doable amount of terms
    Also I have seen this done with complex analysis...might want to wait this one out for TPH or someone more knowledgable in that field

    I have seen this split into two integrals...and you turn the second one into a Laurent series
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  13. #13
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    We can write this in the form of a polylogarithm. You have probably seen the series. \sum_{n=1}^{\infty}\frac{z^{n}}{n^{a}}

    The first term is what Moo came up with:

    \frac{-1}{6}x^{6}+x^{5}ln(1-e^{x})+\sum_{n=1}^{\infty}\frac{(e^{x})^{n}}{n^{2}  }\cdot{5x^{4}}

    -\sum_{n=1}^{\infty}\frac{(e^{x})^{n}}{n^{3}}\cdot{  20x^{3}}+\sum_{n=1}^{\infty}\frac{(e^{x})^{n}}{n^{  4}}\cdot{60x^{2}}

    -\sum_{n=1}^{\infty}\frac{(e^{x})^{n}}{n^{5}}\cdot{  120x}+\sum_{n=1}^{\infty}\frac{(e^{x})^{n}}{n^{6}}  \cdot{120}

    See the pattern?.
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  14. #14
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    Quote Originally Posted by Moo View Post
    What about this :

    I=\int \frac{x^5}{e^x-1} dx

    x \rightarrow -x

    I=\int \frac{x^5}{e^{-x}-1} dx

    Multiply above and below by -e^x :

    I=\int \frac{-x^5 e^x}{e^x-1} dx

    -------------

    2I=\int \frac{x^5}{e^x-1} dx+\int \frac{-x^5 e^x}{e^x-1} dx=\int \frac{x^5 (1-e^x)}{e^x-1}=-\int x^5 dx=-\frac{x^6}{6}

    Therefore, I=-\frac{x^6}{12}


    Is there any mistake ? o.O

    I did a very similar mistake in an integral posted in mathlinks recently. And a nice reader enlightened me with this example as to why it is wrong:

    Quote Originally Posted by kevinatcausa
    The problem's the same thing that's wrong with the following argument.

    \int 2x^3 dx = \int (x^2)( 2x dx) = \int u du = \int x dx = \frac {x^2}{2} + c

    The trouble is that \int u du = \frac {u^2}{2} + c, while \int x dx = \frac {x^2}{2} + c. They may look the same, but they're not equal because u and x are different.
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  15. #15
    Moo
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    Quote Originally Posted by Isomorphism View Post

    I did a very similar mistake in an integral posted in mathlinks recently. And a nice reader enlightened me with this example as to why it is wrong:
    Yes, because when changing variable, either you change the boundaries, either you substitute back at the end. That's why you can't do this way
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