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Math Help - Integrating by parts ?

  1. #1
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    Integrating by parts ?

    \int_{-\infty}^{\infty} \frac{x}{e^x dx}

    The e^x should be e^ modulus x , but i can't seemed to key out the modulus thingy.

    Ok, I tried the by parts method. Letting u = 1/e^mod x , and dv = x dx

    but once i done it , i need to do by parts again. And it goes on and on...
    never ending...

    What did i do to do it uncorrectly ?

    Thanks
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  2. #2
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    I'm not quite sure what you're doing.

    \int x \cdot e^{-x}\;dx\;=\;\int x\;d\left(-e^{-x}\right)\;=\;-x \cdot e^{-x}\;+\;\int e^{-x}\;dx

    Just one application. You're almost done, except that you will have some trouble with convergence on the negative side.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    You have a bit of a problem here.

    Let us just consider this integral on the half-line (-\infty,0)=\mathbb{R}^-

    Then a simple substitution leads us to \int_{\mathbb{R}^-}\frac{x}{e^{-x}}dx=-\int_{\mathbb{R}^+}e^z\cdot z dz which obviously tends to negative infinity.

    and since clearly \int_{\mathbb{R}^+}\frac{x}{e^x}\le\int_{\mathbb{R  }^+}\frac{dx}{x^2} converges we see that

    \int_{\mathbb{R}}\frac{x}{e^{-x}}dx=\int_{\mathbb{R}^-}\frac{x}{e^{-x}}dx+\int_{\mathbb{R}+}\frac{x}{e^x}dx diverges.

    EDIT: Sorry, a little late.
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    You have a bit of a problem here.

    Let us just consider this integral on the half-line (-\infty,0)=\mathbb{R}^-

    Then a simple substitution leads us to \int_{\mathbb{R}^-}\frac{x}{e^{-x}}dx=-\int_{\mathbb{R}^+}e^z\cdot z dz which obviously tends to negative infinity.

    and since clearly \int_{\mathbb{R}^+}\frac{x}{e^x}\le\int_{\mathbb{R  }^+}\frac{dx}{x^2} converges we see that

    \int_{\mathbb{R}}\frac{x}{e^{-x}}dx=\int_{\mathbb{R}^-}\frac{x}{e^{-x}}dx+\int_{\mathbb{R}+}\frac{x}{e^x}dx diverges.

    EDIT: Sorry, a little late.
    However, the answer provided by my lecturer was , it converges to 0. He did not gave us the solution so i have no idea how he did it. All he said was to try integrate by parts.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by xcluded View Post
    However, the answer provided by my lecturer was , it converges to 0. He did not gave us the solution so i have no idea how he did it. All he said was to try integrate by parts.
    - Wolfram|Alpha sweet sweet validation.
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    - Wolfram|Alpha sweet sweet validation.
    Okay. Thanks
    Btw what does the modulus X means to the question ?
    Does it matter to the question if the modulus exist or don't exist?

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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by xcluded View Post
    Okay. Thanks
    Btw what does the modulus X means to the question ?
    Does it matter to the question if the modulus exist or don't exist?

    I'm sorry. What absolute value?
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  8. #8
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    Quote Originally Posted by Drexel28 View Post
    I'm sorry. What absolute value?
    - Wolfram|Alpha

    Because i am wondering what is their difference if the question was changed to this instead.
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by xcluded View Post
    - Wolfram|Alpha

    Because i am wondering what is their difference if the question was changed to this instead.
    Aha! I see what you are saying.

    Once again make note that \int_{\mathbb{R}}=\int_{\mathbb{R}^-}\frac{x}{e^{|x|}}dx+\int_{\mathbb{R}^+}\frac{x}{e  ^{|x|}}dx=I_1+I_2. And we can make a neat observation. In I_1 let x=-z to obtain I_1=-\int_{\infty}^0\frac{-z}{e^{|-z|}}dz=\int_0^{\infty}\frac{-z}{e^{|z|}}dz=-\int_{\mathbb{R}^+}\frac{z}{e^{|z|}}dz=-I_2. So \int_{\mathbb{R}}\frac{x}{e^{|x|}}dx=I_1+I_2=-I_2+I_2=0


    Remark: And if you're really inquisitive \int_{-\infty}^{\infty}\frac{|x|}{e^{|x|}}dx has another similar trick. If we make the same substituiont in I_1 as the previous one we get I_1=I_2 so \int_{\mathbb{R}}\frac{|x|}{e^{|x|}}dx=2\int_{\mat  hbb{R}^+}\frac{|x|}{e^{|x|}}dx=2\int_{\mathbb{R}^+  }\frac{x}{e^x}dx
    Last edited by Drexel28; November 11th 2009 at 10:18 PM.
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  10. #10
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    Thank you drexel28 !

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