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Thread: Integration by parts

  1. #1
    Senior Member Vinod's Avatar
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    Integration by parts

    Hello,
    How to integrate this integral by using integration by parts. $\Gamma(n)=\displaystyle\int_0^1(\ln{\frac1x})^{n-1}dx$

    Someone has answered it as follows:-$\frac{-x}{n}\large(\ln{\frac1x}\large)^n|_{x=0}^{x=1}+ \displaystyle\int_0^1\frac1n \large(\ln{\frac1x}\large)^n=\frac1n \displaystyle\int_0^1\large(\ln{ \frac1x}\large)^n$

    And initial case n=1 $\displaystyle\int_0^1 \large(\ln{\frac1x}\large)^0 dx=1$

    So by induction $\displaystyle\int_0^1 \large(\ln{\frac1x}\large)^{n-1} dx=(n-1)!$

    Now i didn't understand how the integration by parts was done. If anyone knows it,reply it.
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  2. #2
    Senior Member Vinod's Avatar
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    Re: Integration by parts

    Quote Originally Posted by Vinod View Post
    Hello,
    How to integrate this integral by using integration by parts. $\Gamma(n)=\displaystyle\int_0^1(\ln{\frac1x})^{n-1}dx$

    Someone has answered it as follows:$\frac{-x}{n}\large(\ln{\frac1x}\large)^n|_{x=0}^{x=1}+ \displaystyle\int_0^1\frac1n \large(\ln{\frac1x}\large)^n=\frac1n \displaystyle\int_0^1\large(\ln{ \frac1x}\large)^n$

    And initial case n=1 $\displaystyle\int_0^1 \large(\ln{\frac1x}\large)^0 dx=1$

    So by induction $\displaystyle\int_0^1 \large(\ln{\frac1x}\large)^{n-1} dx=(n-1)!$

    Now i didn't understand how the integration by parts was done. If anyone knows it,reply it.
    I have deleted the minus sign before $\frac{-x}{n}$
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  3. #3
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    Re: Integration by parts

    $u=\left( \ln \dfrac 1 x \right)^n, dv=dx$
    Thanks from Vinod
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  4. #4
    Senior Member Vinod's Avatar
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    Re: Integration by parts

    Quote Originally Posted by SlipEternal View Post
    $u=\left( \ln \dfrac 1 x \right)^n, dv=dx$
    Hello,
    Let say $I_{n-1}=\displaystyle\int \left(\ln{\frac1x}\right)^{n-1}$

    $I_{n-1}= \left( \ln{ \frac1x}\right)^{n-1}\displaystyle\int dx -\displaystyle\int\left( \frac{d \left(\ln{\frac1x}\right)^{n-1}}{dx}\displaystyle\int dx \right) dx$

    $I_{n-1}=x\left(\ln{\frac1x}\right)^{n-1}-(n-1)\displaystyle\int \left(\ln{\frac1x}\right)^{n-2} dx$

    So $I_{n-1}=x\left(\ln{\frac1x}\right)^{n-1}-(n-1)I_{n-2}$

    But the answer posted by other fellow is something different in post#1. You are also saying u=\left( \ln\frac1x \right)^n$. How it is? I didn't follow.
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  5. #5
    Senior Member Vinod's Avatar
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    Re: Integration by parts

    Quote Originally Posted by Vinod View Post
    Hello,
    Let say $I_{n-1}=\displaystyle\int \left(\ln{\frac1x}\right)^{n-1} dx$

    $I_{n-1}= \left( \ln{ \frac1x}\right)^{n-1}\displaystyle\int dx -\displaystyle\int\left( \frac{d \left(\ln{\frac1x}\right)^{n-1}}{dx}\displaystyle\int dx \right) dx$

    $I_{n-1}=x\left(\ln{\frac1x}\right)^{n-1}-(n-1)\displaystyle\int \left(\ln{\frac1x}\right)^{n-2} dx$

    So $I_{n-1}=x\left(\ln{\frac1x}\right)^{n-1}-(n-1)I_{n-2}$

    But the answer posted by other fellow is something different in post#1. You are also saying u=\left( \ln\frac1x \right)^n$. How it is? I didn't follow.
    Hello,
    Please read $I_{n-1}=\displaystyle\int \left(\ln {\frac1x}\right)^{n-1} dx$.
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  6. #6
    Senior Member Vinod's Avatar
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    Re: Integration by parts

    Quote Originally Posted by Vinod View Post
    Hello,
    Please read $I_{n-1}=\displaystyle\int \left(\ln {\frac1x}\right)^{n-1} dx$.
    Hello,
    Alternatively, we can also say $I_{n-1}=\frac{x}{n} \left(\ln{\frac1x}\right)^n-\frac1nI_n$
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  7. #7
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    Re: Integration by parts

    I gave you the integration by parts that gives the result in your first post. I did not say it had anything to do with your original problem in your first post. I am not sure I understand what you are saying in any of the following posts. You may be making a mistake with the following differentiation.

    If

    $$u=\left( \ln \dfrac 1 x \right)^{n-1}$$

    Find $du$
    Last edited by SlipEternal; Oct 14th 2018 at 09:18 AM.
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  8. #8
    Senior Member Vinod's Avatar
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    Re: Integration by parts

    Quote Originally Posted by SlipEternal View Post
    I gave you the integration by parts that gives the result in your first post. I did not say it had anything to do with your original problem in your first post. I am not sure I understand what you are saying in any of the following posts. You may be making a mistake with the following differentiation.

    If

    $$u=\left( \ln \dfrac 1 x \right)^{n-1}$$

    Find $du$
    Hello,
    $du= \frac{n-1}{x} \left( \ln{\frac1x} \right)^{n-2} dx$
    Last edited by Vinod; Oct 14th 2018 at 10:14 AM.
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  9. #9
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    Re: Integration by parts

    Quote Originally Posted by Vinod View Post
    Hello,
    $du= \frac{n-1}{x}\displaystyle\int \left( \ln{\frac1x} \right)^{n-2} dx$
    No. That is not how differentiation works. There should be no integral in $du$.
    Thanks from Vinod
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  10. #10
    Senior Member Vinod's Avatar
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    Re: Integration by parts

    Quote Originally Posted by SlipEternal View Post
    No. That is not how differentiation works. There should be no integral in $du$.
    Hello, I have edited post#8. Look at that post. Whatever i posted in other posts are reduction formulas
    Last edited by Vinod; Oct 14th 2018 at 10:26 AM.
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  11. #11
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    Re: Integration by parts

    Quote Originally Posted by Vinod View Post
    Hello, I have edited post#8. Look at that post. Whatever i posted in other posts are reduction formulas
    You missed a negative sign.

    $$du = -\dfrac{(n-1)}{x}\left( \ln \dfrac 1 x \right)^{n-2}dx$$

    So,

    $$\int_0^1 \left( \ln \dfrac 1 x \right)^{n-1}dx = \left. x\left( \ln \dfrac 1 x \right)^{n-1}\right]_0^1 + (n-1)\int_0^1\left( \ln \dfrac 1 x \right)^{n-2}dx$$

    When you plug in 1, you get $\ln \dfrac 1 x =0$. As $x\to 0$, you get 0 times a number which is still zero. So, that first term is zero. You just reduced the exponent by one.
    Thanks from topsquark and Vinod
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