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Math Help - [SOLVED] Integration

  1. #1
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    [SOLVED] Integration

    Why is:

    \frac{1}{2\sigma}\int_{-\infty}^{\infty}xexp(-\frac{|x|}{\sigma}).dx=0

    \frac{1}{2\sigma}\int_{-\infty}^{\infty}x^2exp(-\frac{|x|}{\sigma}).dx=\frac{1}{\sigma}\int_{0}^{\  infty}x^2exp(-\frac{x}{\sigma}).dx
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  2. #2
    JML
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    It's because the first integral contains an odd function and is therefore zero as \int_{-\infty}^{\infty} f(x) dx=0 if f(x) is odd.
    The second integral contains an even function so \int_{-\infty}^{\infty} f(x) dx=2 \int_{0}^{\infty}f(x) dx if f(x) is even.

    also it should be noted that f(x) =  x exp(-\frac{|x|}{\sigma}) contains no asymptotes over the domain of integration so the above holds.
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  3. #3
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    Quote Originally Posted by JML View Post
    It's because the first integral contains an odd function and is therefore zero as \int_{-\infty}^{\infty} f(x) dx=0 if f(x) is odd.
    [snip]
    Although the result is true in this case, the argument is not valid.

    For example, the same argument 'proves' that \int_{- \infty}^{+ \infty} \frac{x}{1 + x^2} \, dx = 0 (the Cauchy Principle value). However, a more rigorous analysis proves that this integral does not in fact exist. (You can research this yourself by researching the fact that the mean of a Cauchy distribution does not exist).
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  4. #4
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    As long as f is an odd function (f(-x)= -f(x)) and the integral exists, \int_{-a}^a f(x) dx= 0.

    \int_{-a}^a f(x) dx= \int_{-a}^0 f(x) dx+ \int_0^a f(x) dx
    In the first integral, let u= -x so du= -dx and x= -a gives u= a. The integral becomes \int_a^0 f(-u)(-du)= \int_0^a f(-u)du = \int_0^a f(u)du= -\int_0^a f(u)du so \int_{-a}^a f(x) dx= \int_0^a f(x)dx- \int_0^a f(u)du= 0 since the "x" and "u" are just dummy variables.

    With f an even function, f(-x)= f(x), you can do the same thing and get "+" between the integrals rather than "-". \int_{-a}^a f(x)dx= \int_0^a f(x)dx+ \int_0^a f(u)du= 2\int_0^a f(x)dx.

    Of course, for that to work all those integrals have to exist.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    As long as f is an odd function (f(-x)= -f(x)) and the integral exists, \int_{-a}^a f(x) dx= 0.

    \int_{-a}^a f(x) dx= \int_{-a}^0 f(x) dx+ \int_0^a f(x) dx
    In the first integral, let u= -x so du= -dx and x= -a gives u= a. The integral becomes \int_a^0 f(-u)(-du)= \int_0^a f(-u)du = \int_0^a f(u)du= -\int_0^a f(u)du so \int_{-a}^a f(x) dx= \int_0^a f(x)dx- \int_0^a f(u)du= 0 since the "x" and "u" are just dummy variables.

    With f an even function, f(-x)= f(x), you can do the same thing and get "+" between the integrals rather than "-". \int_{-a}^a f(x)dx= \int_0^a f(x)dx+ \int_0^a f(u)du= 2\int_0^a f(x)dx.

    Of course, for that to work all those integrals have to exist.
    Yes, I agree with all this.

    But the point that needs to be noted is that if f(x) is odd, \int_{- \infty}^{+ \infty} f(x) \, dx does not necessarily exist (although the Cauchy Principle value exists and is equal to zero).
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