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  1. #1
    Super Member Random Variable's Avatar
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    integrals

    Don't hide your solutions.


    1)  \int_{0}^{\infty} \frac{\sin^{2} (ax)}{x^{2}(1+x^{2})} \ dx \ \ a>0


    2)  \int_{0}^{\infty} \frac{e^{-x^{2}}}{\big(x^{2}+ \frac{1}{2}\big)^{2}} \ dx


    3)  \int_{0}^{\infty} \frac{x^{n}}{1+\cosh x} \ dx \ \ n \in \mathbb{N}, \ n > 1
    Last edited by Random Variable; October 8th 2011 at 08:05 PM.
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  2. #2
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    Re: integrals

    Nice problems!

    (1) can be written as \int_0^\infty \frac{\sin^2 (ax)}{x^2}dx-\int_0^\infty \frac{\sin^2 (ax)}{1+x^2}

    The first integral is known and its value is \frac{\pi}{2}a. The second can be calculated with contour integration, yielding \frac{\pi}{4}\left[1-e^{-2a}\right].





    (3) note that \frac{1}{1+\cosh x} = \frac{2}{e^x+2+e^{-x}}=\frac{2}{(e^{\frac{1}{2}x}+e^{-\frac{1}{2}x})^2}=\frac{2e^{-x}}{(1+e^{-x})^2}

    and since \frac{t}{(1-t)^2}=\sum_{k=1}^\infty kt^k for |t|<1, we can put t=-e^{-x} to obtain:

    \frac{1}{1+\cosh x} = 2 \sum_{k=1}^\infty (-1)^{k+1} ke^{-kx}

    Also note that \int_0^\infty x^n e^{-kx}dx = \frac{\Gamma(n+1)}{k^{n+1}} and \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^n} = \left(1-2^{1-n}\right)\zeta(n).

    From all these we obtain:

    \int_0^\infty \frac{x^n}{1+\cosh x} dx = 2 \int_0^\infty \left[\sum_{k=1}^\infty (-1)^{k+1}ke^{-kx}\right]dx

    = 2 \sum_{k=1}^\infty (-1)^{k+1}k \int_0^\infty x^ne^{-kx}\right]dx = 2 \sum_{k=1}^\infty (-1)^{k+1} \frac{\Gamma(n+1)}{k^n}

    =2\left(1-2^{1-n}\right)\zeta(n)\Gamma(n+1)

    and when n>1 is a natural number, this is 2\left(1-2^{1-n}\right)n!\zeta(n)





    (2) Let J(a,b)=\int_0^\infty \frac{e^{-bx^2}}{x^2+a^2} dx, with a,b>0.

    Then \int_0^\infty \frac{e^{-bx^2}}{(x^2+a^2)^2}dx = -\frac{1}{2a}\frac{\partial J}{\partial a}, and we're looking for this integral, evaluated when a^2=\frac{1}{2}, b=1.

    To calculate J, note that:

    a^2 J - \frac{\partial J}{\partial b} = \int_0^\infty \frac{a^2 e^{-bx^2}}{x^2+a^2}dx+\int_0^\infty \frac{x^2 e^{-bx^2}}{x^2+a^2}dx = \int_0^\infty e^{-bx^2}dx = \frac{1}{2}\sqrt\frac{\pi}{b}

    So that:

    \frac{\partial}{\partial b}\left(e^{-a^2b} J\right) = -\frac{1}{2}\sqrt\frac{\pi}{b} e^{-a^2b}

    Now let f(t) = \int_0^t e^{-x^2}dx.

    It is easily checked that the indefinite integral with respect to b of the right hand side of the previous equation is -\frac{\sqrt\pi}{a} f(a\sqrt{b})+C, so:

    e^{-a^2b} J = -\frac{\sqrt\pi}{a} f(a\sqrt{b})+C(a)

    From the definition of J it is obvious that the left hand side vanishes when b=\infty, so the right hand side also vanishes. Thus, using f(\infty)=\frac{\sqrt\pi}{2}, we get C=\frac{\pi}{2a}. Therefore:

    J=\frac{\sqrt\pi e^{a^2 b}}{a}\left[\frac{\sqrt\pi}{2}-f(a\sqrt{b})\right]

    Now we differentiate with respect to a, remembering that \frac{d}{dt} f(t) = e^{-t^2}:

    \frac{\partial J}{\partial a} = \sqrt{\pi}e^{a^2 b}\frac{2a^2b-1}{a^2} \left[\frac{\sqrt\pi}{2}-f(a\sqrt{b})\right]-\frac{\sqrt{\pi b}}{a}.

    Now, as we noted, \int_0^\infty \frac{e^{-bx^2}}{(x^2+a^2)^2}dx = -\frac{1}{2a}\frac{\partial J}{\partial a}, so finally we have:

    \int_0^\infty \frac{e^{-bx^2}}{(x^2+a^2)^2}dx = \frac{\sqrt{\pi b}}{2a^2}-\sqrt{\pi}e^{a^2 b}\frac{2a^2b-1}{2a^3} \left[\frac{\sqrt\pi}{2}-f(a\sqrt{b})\right].

    Fortunately, when a^2=\frac{1}{2} and b=1 the second term vanishes, and thus

    \int_0^\infty \frac{e^{-x^2}}{(x^2+\frac{1}{2})^2}dx = \sqrt{\pi}
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  3. #3
    Super Member Random Variable's Avatar
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    Re: integrals

    Surprisingly, problem 2 can also be solved by integrating by parts.


     \int_{0}^{\infty} \frac{e^{-x^{2}}}{(x^2 + \frac{1}{2})^{2}}} \ dx = \int_{0}^{\infty} \frac{e^{-x^{2}}}{x} \frac{x}{(x^{2}+\frac{1}{2})^{2}}} \ dx


     = -\frac{1}{2} \frac{e^{-x^{2}}}{x(x^{2}+\frac{1}{2})^{2}} \Big|^{\infty}_{0} - \int_{0}^{\infty} \frac{e^{-x^{2}}}{x^{2}} \ dx


     = \Big(-\frac{1}{2} \frac{e^{-x^{2}}}{x(x^{2}+\frac{1}{2})^{2}} + \frac{e^{-x^{2}}}{x} \Big) \Big|_{0}^{\infty} + 2 \int_{0}^{\infty} e^{-x^{2}} \ dx


     = \frac{xe^{-x^{2}}}{x^{2}+\frac{1}{2}} \Big|^{\infty}_{0} + 2 \int_{0}^{\infty} e^{-x^{2}} = \sqrt{\pi}
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  4. #4
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    Re: integrals

    This another Solution :

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