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Math Help - Can you check this integration for me please

  1. #1
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    Can you check this integration for me please



    I think the denominator should be the other way around: it-lambda

    ip means integration by parts
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  2. #2
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    Quote Originally Posted by adam_leeds View Post


    I think the denominator should be the other way around: it-lambda

    ip means integration by parts
    If you are trying to evaluate \lambda ^2 \int_0^{\infty}{x\,e^{x(i\,t - \lambda)}\,dx}


    \lambda ^2 \int_0^{\infty}{x\,e^{x(i\,t - \lambda)}\,dx} = \frac{\lambda ^2}{i\,t - \lambda}\int_0^{\infty}{x(i\,t - \lambda)e^{x(i\,t - \lambda)}\,dx}

    Now using integration by parts with

    u = x so that du = 1

    and dv = (i\,t - \lambda)e^{x(i\,t - \lambda)} so that v = e^{x(i\,t - \lambda)}

    we find

    \frac{\lambda ^2}{i\,t - \lambda}\int_0^{\infty}{x(i\,t -  \lambda)e^{x(i\,t - \lambda)}\,dx} = \left(\frac{\lambda ^2}{i\,t - \lambda}\right)\lim_{\varepsilon \to \infty}\left[x\,e^{x(i\,t - \lambda)} - \int{e^{x(i\,t - \lambda)}\,dx}\right]_0^{\varepsilon}

    =\left(\frac{\lambda ^2}{i\,t - \lambda}\right)\lim_{\varepsilon \to \infty}\left[x\,e^{x(i\,t - \lambda)} - \frac{e^{x(i\,t - \lambda)}}{i\,t - \lambda}\right]_0^{\varepsilon}

    Can you finish?
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  3. #3
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    Quote Originally Posted by Prove It View Post
    If you are trying to evaluate \lambda ^2 \int_0^{\infty}{x\,e^{x(i\,t - \lambda)}\,dx}


    \lambda ^2 \int_0^{\infty}{x\,e^{x(i\,t - \lambda)}\,dx} = \frac{\lambda ^2}{i\,t - \lambda}\int_0^{\infty}{x(i\,t - \lambda)e^{x(i\,t - \lambda)}\,dx}

    Now using integration by parts with

    u = x so that du = 1

    and dv = (i\,t - \lambda)e^{x(i\,t - \lambda)} so that v = e^{x(i\,t - \lambda)}

    we find

    \frac{\lambda ^2}{i\,t - \lambda}\int_0^{\infty}{x(i\,t - \lambda)e^{x(i\,t - \lambda)}\,dx} = \left(\frac{\lambda ^2}{i\,t - \lambda}\right)\lim_{\varepsilon \to \infty}\left[x\,e^{x(i\,t - \lambda)} - \int{e^{x(i\,t - \lambda)}\,dx}\right]_0^{\varepsilon}

    =\left(\frac{\lambda ^2}{i\,t - \lambda}\right)\lim_{\varepsilon \to \infty}\left[x\,e^{x(i\,t - \lambda)} - \frac{e^{x(i\,t - \lambda)}}{i\,t - \lambda}\right]_0^{\varepsilon}

    Can you finish?
    Thanks i don't know how to use latex so paint will have to do lol

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  4. #4
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    Im still sure the op is wrong any help?
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  5. #5
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    Notice that (a-b)^2 = (b-a)^2 for any a,b.

    So \frac{\lambda^2}{(\lambda - i t)^2} = \frac{\lambda^2}{(it-\lambda)^2}

    If you need convincing, just try to square both of the denominators and see that you get the same result.
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  6. #6
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    Quote Originally Posted by drumist View Post
    Notice that (a-b)^2 = (b-a)^2 for any a,b.

    So \frac{\lambda^2}{(\lambda - i t)^2} = \frac{\lambda^2}{(it-\lambda)^2}

    If you need convincing, just try to square both of the denominators and see that you get the same result.
    Thank you
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