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Definite integral

  1. Jan 24, 2006 #1
    Can anyone help me integrating the folowwing integrand from zero to infinity:
    where a,b and c are real and pisitive constants.
    Last edited: Jan 24, 2006
  2. jcsd
  3. Jan 24, 2006 #2
    First term diverges with [tex]x \rightarrow \infty[/tex]. Is the region really from 0 to infinity? If possible, please post in TeX.
    But first term is not difficult. Using partial integral,

    [tex]\int x (\ln ax)^2 dx = 1/2 x^2 (\ln ax)^2 - \int \frac {ax^2} {ax} \ln ax dx = \frac 1 2 x^2 (\ln ax)^2 - \int x \ln ax dx[/tex].

    Using partial integral on the second term, it is

    [tex]\frac 1 2 x^2 \ln ax - \frac 1 2 \int ax^2 \frac 1 {ax} dx = \frac 1 2 x^2 \ln ax - \frac 1 4 x^2[/tex]

    So in total,

    [tex]\int x (\ln ax)^2 dx = \frac 1 2 x^2((\ln ax))^2 - \ln ax) + \frac 1 4 x^2[/tex].

    The second term can be integrated from [tex]-\infty \rightarrow \infty[/tex]
    Last edited: Jan 24, 2006
  4. Jan 24, 2006 #3
    As for [tex]\exp(-bx^2+cx)[/tex], it can be integrated for some kinds of limited regions...(AFAIK) using Gaussian integral. i.e. Put

    [tex]I = \int_{-\infty}^\infty \exp(-x^2) dx[/tex].


    [tex]I^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty \exp(-(x^2 + y^2)) dx dy = 2 \pi \int_0^\infty r \exp(-r^2) dr = \pi[/tex]

    So [tex]I = \sqrt \pi.[/tex]

    But exp(-bx^2+cx) is another thing if the range is different.... If the region is from [tex]-\infty \rightarrow \infty[/tex]

    [tex]\int_{-\infty}^\infty \exp(-bx^2 + cx) dx = \int_{\infty}^\infty \exp(-b(x-\frac c {2b}) ^ 2 + \frac {c^2} {4b}) dx = \sqrt \pi \exp (c^2/4b) / \sqrt b[/tex]

    I cannot go further..:frown:
    Last edited: Jan 24, 2006
  5. Jan 24, 2006 #4
    I'm very sorry. I was mistaken. It should be a multiplication of the two terms, not sum. The integrand is now corrected in the original message.
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