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Math Help - What Am I Doing Wrong? - Integral

  1. #1
    Super Member angel.white's Avatar
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    What Am I Doing Wrong? - Integral

    Back of the book says the answer is zero, but I keep getting infinity (which I think means it is divergent), and I don't know where the discrepancy lies :/

    \int_{-\infty}^\infty xe^{-x^2}

    =\lim_{N\to\infty}\int_0^N xe^{-x^2} + \lim_{B\to\infty}\int_{-B}^0 xe^{-x^2}

    =\lim_{N\to\infty}\left[\frac {-1}2e^{-x^2}\right]_0^N + \lim_{B\to\infty}\left[\frac {-1}2e^{-x^2}\right]_{-B}^0


    =\lim_{N\to\infty}\left[\frac {-1}{2e^{x^2}}\right]_0^N + \lim_{B\to\infty}\left[\frac {-1}{2e^{x^2}}\right]_{-B}^0

    =\lim_{N\to\infty}\left[\frac{-1}{2e^{N^2}}+\frac{1}{2e^{0^2}}\right] +\lim_{B\to\infty}\left[\frac {-1}{2e^{0^2}}+\frac {1}{2e^{-B^2}}\right]

    =\lim_{N\to\infty}\left[\frac{-1}{2e^{N^2}}+\frac{1}{2}\right] +\lim_{B\to\infty}\left[\frac {-1}{2}+\frac {e^{B^2}}{2}\right]

    =\left[\frac{-1}{2e^{\infty^2}}+\frac{1}{2}\right] +\left[\frac {-1}{2}+\frac {e^{\infty^2}}{2}\right]

    =\frac{-1}{\infty}+\frac{1}{2} + \frac {-1}{2}+\frac {\infty}{2}

    =\frac{-1}{\infty}+\frac {\infty}{2}

    =0+\infty

    =\infty
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    Quote Originally Posted by angel.white View Post
    Back of the book says the answer is zero, but I keep getting infinity (which I think means it is divergent), and I don't know where the discrepancy lies :/

    \int_{-\infty}^\infty xe^{-x^2}

    =\lim_{N\to\infty}\int_0^N xe^{-x^2} + \lim_{B\to\infty}\int_{-B}^0 xe^{-x^2}

    =\lim_{N\to\infty}\left[\frac {-1}2e^{-x^2}\right]_0^N + \lim_{B\to\infty}\left[\frac {-1}2e^{-x^2}\right]_{-B}^0


    =\lim_{N\to\infty}\left[\frac {-1}{2e^{x^2}}\right]_0^N + \lim_{B\to\infty}\left[\frac {-1}{2e^{x^2}}\right]_{-B}^0

    =\lim_{N\to\infty}\left[\frac{-1}{2e^{N^2}}+\frac{1}{2e^{0^2}}\right] +\lim_{B\to\infty}\left[\frac {-1}{2e^{0^2}}+\frac {1}{2e^{-B^2}}\right]

    =\lim_{N\to\infty}\left[\frac{-1}{2e^{N^2}}+\frac{1}{2}\right] +\lim_{B\to\infty}\left[\frac {-1}{2}+\frac {e^{B^2}}{2}\right]

    =\left[\frac{-1}{2e^{\infty^2}}+\frac{1}{2}\right] +\left[\frac {-1}{2}+\frac {e^{\infty^2}}{2}\right]

    =\frac{-1}{\infty}+\frac{1}{2} + \frac {-1}{2}+\frac {\infty}{2}

    =\frac{-1}{\infty}+\frac {\infty}{2}

    =0+\infty

    =\infty
    You're integrating an odd function. Letting the integral limits be x = -L and x = L, the integral will equal zero. And of course you can then take the limit L --> oo and get zero .....
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  3. #3
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    Quote Originally Posted by angel.white View Post
    Back of the book says the answer is zero, but I keep getting infinity (which I think means it is divergent), and I don't know where the discrepancy lies :/

    \int_{-\infty}^\infty xe^{-x^2}

    =\lim_{N\to\infty}\int_0^N xe^{-x^2} + \lim_{B\to\infty}\int_{-B}^0 xe^{-x^2}

    =\lim_{N\to\infty}\left[\frac {-1}2e^{-x^2}\right]_0^N + \lim_{B\to\infty}\left[\frac {-1}2e^{-x^2}\right]_{-B}^0


    =\lim_{N\to\infty}\left[\frac {-1}{2e^{x^2}}\right]_0^N + \lim_{B\to\infty}\left[\frac {-1}{2e^{x^2}}\right]_{-B}^0

    =\lim_{N\to\infty}\left[\frac{-1}{2e^{N^2}}+\frac{1}{2e^{0^2}}\right] +\lim_{B\to\infty}\left[\frac {-1}{2e^{0^2}}+\frac {1}{2e^{-B^2}}\right] Mr F says: And the mistake is in this line ..... When you sub \, x = -B \, into \, x^2\, , the result is \, B^2\, , not \, -B^2\, .....

    =\lim_{N\to\infty}\left[\frac{-1}{2e^{N^2}}+\frac{1}{2}\right] +\lim_{B\to\infty}\left[\frac {-1}{2}+\frac {e^{B^2}}{2}\right]

    =\left[\frac{-1}{2e^{\infty^2}}+\frac{1}{2}\right] +\left[\frac {-1}{2}+\frac {e^{\infty^2}}{2}\right]

    =\frac{-1}{\infty}+\frac{1}{2} + \frac {-1}{2}+\frac {\infty}{2}

    =\frac{-1}{\infty}+\frac {\infty}{2}

    =0+\infty

    =\infty
    ..
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  4. #4
    Super Member angel.white's Avatar
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    Thank you, glad it was something stupid and not a complete lack of understanding on how to do these.

    Pretty confident, then, that I can do these now, but I already uploaded my next two problems, so if someone wouldn't mind checking the answers, I would appreciate it. They're even problems, so not in the back of the book, and the calculator I usually use to check my work doesn't seem to like infinite boundaries.


    ----------------
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    Quote Originally Posted by angel.white View Post
    Thank you, glad it was something stupid and not a complete lack of understanding on how to do these.

    Pretty confident, then, that I can do these now, but I already uploaded my next two problems, so if someone wouldn't mind checking the answers, I would appreciate it. They're even problems, so not in the back of the book, and the calculator I usually use to check my work doesn't seem to like infinite boundaries.


    [snip]
    Instant feedback: The first one is wrong I'm sorry to say. Note that ln(0) is undefined, NOT 1 ..... There's a couple of other things wrong too ... You can't multiply oo by 2 ....!! Under the change of variable a = 2x - 5, x = -oo => a = -oo.
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    Quote Originally Posted by angel.white View Post
    Thank you, glad it was something stupid and not a complete lack of understanding on how to do these.

    Pretty confident, then, that I can do these now, but I already uploaded my next two problems, so if someone wouldn't mind checking the answers, I would appreciate it. They're even problems, so not in the back of the book, and the calculator I usually use to check my work doesn't seem to like infinite boundaries.

    [snip]

    Your answer is correct, but again the error of treating oo like a number. Under the change of variable a = x^2 + 2, x = +oo => a = +oo .....
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  7. #7
    Super Member angel.white's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Instant feedback: The first one is wrong I'm sorry to say. Note that ln(0) is undefined, NOT 1
    >.< not my night, think I'll pick these back up tomorrow when my mind is fresher.
    Quote Originally Posted by mr fantastic View Post
    ..... There's a couple of other things wrong too ... You can't multiply oo by 2 ....!! Under the change of variable a = 2x - 5, x = -oo => a = -oo.
    Okay, how do I remedy that? Put the limit in place pior to changing the limits?

    It's so hard to force myself to do these, I know what I want to do, I have like the next 3 or 4 steps planned in my mind, and then I get bogged down writing out the notation.

    Thank you for your help, btw.
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  8. #8
    Senior Member Peritus's Avatar
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    there's another nice substitution for your second problem, hope you'll find it helpful:


    <br />
\begin{gathered}<br />
  \int_0^\infty  {\frac{x}<br />
{{\left( {x^2  + 2} \right)^2 }}dx}  \hfill \\<br />
   \hfill \\<br />
  x = \sqrt 2 \tan u \hfill \\<br />
  dx = \frac{{\sqrt 2 }}<br />
{{\cos ^2 u}}du \hfill \\ <br />
\end{gathered} <br />

    the new limits are:

    <br />
\begin{gathered}<br />
  0 < \sqrt 2 \tan u < \infty  \hfill \\<br />
  0 < u < \frac{\pi }<br />
{2} \hfill \\ <br />
\end{gathered}


    \int_0^{\frac{\pi }<br />
{2}} {\frac{{\sqrt 2 \tan u}}<br />
{{\left( {2\tan ^2 u + 2} \right)^2 }}\frac{{\sqrt 2 }}<br />
{{\cos ^2 u}}du}  = \frac{1}<br />
{2}\int_0^{\frac{\pi }<br />
{2}} {\sin u\cos udu}  = \frac{1}<br />
{4}\int_0^{\frac{\pi }<br />
{2}} {\sin 2udu}  =

    <br />
 = \left. { - \frac{1}<br />
{8}\cos 2u} \right|_0^{\frac{\pi }<br />
{2}}  = \frac{1}<br />
{4}<br />
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  9. #9
    Super Member angel.white's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Your answer is correct, but again the error of treating oo like a number. Under the change of variable a = x^2 + 2, x = +oo => a = +oo .....
    Okay, I redid the problem, trying to use all limits until the end when I couldn't simplify any more without plugging infinity back into the problem. I am hoping that this fixes the problem of "treating oo like a number." Can someone let me know if this is the correct way to handle this type of problem?

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  10. #10
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    Quote Originally Posted by angel.white View Post
    Okay, I redid the problem, trying to use all limits until the end when I couldn't simplify any more without plugging infinity back into the problem. I am hoping that this fixes the problem of "treating oo like a number." Can someone let me know if this is the correct way to handle this type of problem?

    Looks much better.

    This is not a criticism, but I'd be inclined to do it the following way:

    The substitution a = 2x - 5 means that the integral limits change from -\infty < x < 0 to -\infty < a < -5.

    So the integral becomes \frac{1}{2} \int_{-\infty}^{-5} \frac{1}{a} \, da.

    Then I'd do the limit business on this improper integral ..... I just find it makes life a bit easier.

    But there is nothing wrong with your approach.
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