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Math Help - Techniques of integration (5)

  1. #1
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    Techniques of integration (5)

    \int \left(x \ln x - \frac{(n-1)!}{x^n} \right)e^x \ dx = ?
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  2. #2
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    Quote Originally Posted by NonCommAlg View Post
    \int \left(x \ln x - \frac{(n-1)!}{x^n} \right)e^x \ dx = ?
    I bet there is a better way , but I thought this was fun.

    First lets break this up a bit

     <br />
\int x\ln(x)e^xdx-(n-1)!\int x^{-n}e^xdx<br />

    Lets focus on the 2nd integral to start

    (n-1)!\int x^{-n}e^xdx

    so if we integrate by parts with u=e^x \implies du =e^xdx and dv=x^{-n} \implies v=-\frac{1}{n-1}x^{-n+1}

    (n-1)!\int x^{-n}e^xdx=(n-1)!\left(-\frac{1}{n-1}x^{-n+1}e^x+\frac{1}{n-1}\int x^{-n+1}e^xdx \right)=
    -(n-2)!x^{-n+1}e^{x}+(n-2)!\int x^{-n+1}e^x

    If we integrate by parts again we get

    -(n-2)!x^{-n+1}e^{x}-(n-3)!x^{-n+2}e^{x}+(n+3)!\int x^{-n+2}e^x

    Doing this n-1 times we get

    (n-1)!\int x^{-n}e^xdx=-e^{x}\sum_{i=1}^{n-1}[n-(1+i)]!x^{-n+i}+\int \frac{1}{x}e^{x}dx

    Now lets look at the last bit

    \int \frac{1}{x}e^{x}dx if we integrate this by parts we get

     \int \frac{1}{x}e^{x}dx=e^{x}\ln(x)-\int e^{x}\ln(x)dx

    Now lets integrate the last integral by parts again with

    u=e^{x} \implies du=e^{x},dv=\ln(x ) \implies v=x\ln(x)-x

     \int \frac{1}{x}e^{x}dx=e^{x}\ln(x)-[x\ln(x)-x]e^{x}+\int [x\ln(x)-x]e^xdx

     \int \frac{1}{x}e^{x}dx=e^{x}\ln(x)-[x\ln(x)-x]e^{x}+\int x\ln(x)e^{x}dx-\int xe^xdx

    So finally we get

    (n-1)!\int x^{-n}e^xdx=-e^{x}\sum_{i=1}^{n-1}[n-(1+i)]!x^{-n+i}+e^{x}\ln(x)-[x\ln(x)-x]e^{x}+
    \int x\ln(x)e^{x}dx-\int xe^xdx

    Now subbing this back into the very first line we get

    e^{x}\sum_{i=1}^{n-1}[n-(1+i)]!x^{-n+i}+e^{x}\ln(x)+[x\ln(x)-x]e^{x}+\int xe^xdx

    evaluateing the last integral and simplifying gives

    e^{x}\sum_{i=1}^{n-1}[n-(1+i)]!x^{-n+i}+e^{x}\ln(x)+xe^{x}\ln(x)-e^x
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  3. #3
    Moo
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    Hello,

    Here is what I tried...

    I_n(x)=\int\left(x\ln(x)-\frac{(n-1)!}{x^n}\right)e^x ~dx

    By differentiating this (under the integral sign), we can get :
    I_n'(x)=I_n(x)+\int\left(1+\ln(x)+\frac{n!}{x^{n+1  }}\right)e^x~dx

    I_n'(x)=I_n(x)-I_{n+1}(x)+\int\left(1+\ln(x)+x\ln(x)\right)e^x~dx

    But the latter is exactly x\ln(x)e^x+C

    And we also have I_n'(x)=\left(x\ln(x)-\frac{(n-1)!}{x^n}\right)e^x

    So finally :
    I_{n+1}(x)-I_n(x)=\frac{(n-1)!}{x^n}\cdot e^x+C

    And :
    I_n(x)=e^x \sum_{k=2}^n \frac{(k-2)!}{x^{k-1}}+I_2+D


    That's where I've gotten to. I didn't have much time this morning. And since I'm not at home, I don't have a paper in front of me and cannot continue immediately... I think the problem will be to compute I_2... The sum is just the (n-1) (or (n-2)) th derivative of x/(x-1)=1+1/(x-1), which can easily be found.

    I'll think about it later. See ya
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  4. #4
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    Quote Originally Posted by Moo View Post
    Hello,

    Here is what I tried...

    I_n(x)=\int\left(x\ln(x)-\frac{(n-1)!}{x^n}\right)e^x ~dx

    By differentiating this (under the integral sign), we can get :
    I_n'(x)=I_n(x)+\int\left(1+\ln(x)+\frac{n!}{x^{n+1  }}\right)e^x~dx
    here Moo actually used by parts: x \ln x - \frac{(n-1)!}{x^n} = u, \ e^x dx = dv and the fact that uv=I_n'(x).


    I_n'(x)=I_n(x)-I_{n+1}(x)+\int\left(1+\ln(x)+x\ln(x)\right)e^x~dx

    But the latter is exactly x\ln(x)e^x+C

    And we also have I_n'(x)=\left(x\ln(x)-\frac{(n-1)!}{x^n}\right)e^x

    So finally :
    I_{n+1}(x)-I_n(x)=\frac{(n-1)!}{x^n}\cdot e^x+C

    And :
    I_n(x)=e^x \sum_{k=2}^n \frac{(k-2)!}{x^{k-1}}+I_2+D


    That's where I've gotten to. I didn't have much time this morning. And since I'm not at home, I don't have a paper in front of me and cannot continue immediately... I think the problem will be to compute I_2... The sum is just the (n-1) (or (n-2)) th derivative of x/(x-1)=1+1/(x-1), which can easily be found.

    I'll think about it later. See ya
    this is a nice approach. you probably don't need to think about finding I_2 because \emptyset's solution, which is exactly how i created the problem, has already answered your question.
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