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Math Help - Find integral for this

  1. #1
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    Find integral for this

    Hi i'm lookin for the integral of t*exp(-abs(t))sin(wt) dt with limits at +infinity and -infinity

    Anyone available to help?
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  2. #2
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    Well, I have a solution but no steps..

    I hope this will help you though.

     \frac{4w}{(1+w^2)^2}
    Greets,Leslon
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by bcvw85 View Post
    Hi i'm lookin for the integral of t*exp(-abs(t))sin(wt) dt with limits at +infinity and -infinity
    First, let's get rid off the absolute value. As t\mapsto t\sin(\omega t)\exp(-|t|) is an even function,

    I=\int_{-\infty}^{\infty}t\sin(\omega t)\exp(-|t|)\,\mathrm{d}t=2\int_0^{\infty}t\sin(\omega t)\exp(-t)\,\mathrm{d}t

    Then, one can use the expression of the sine in terms of exponential functions : \sin(\omega t)=\frac{\exp(i\omega t)-\exp(-i\omega t)}{2i} which gives

    \begin{aligned} I&=\int_0^{\infty}t\left(\frac{\exp(i\omega t)-\exp(-i\omega t)}{2i}\right)\exp(-t)\,\mathrm{d}t\\ &=\int_0^{\infty}t\left(\frac{\exp((i\omega-1) t)-\exp(-(1+i\omega) t)}{2i}\right)\,\mathrm{d}t\end{aligned}

    Using integration by parts one can get rid off the t which is outside the exponential functions.
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  4. #4
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    Quote Originally Posted by flyingsquirrel View Post
    Hi


    First, let's get rid off the absolute value. As t\mapsto t\sin(\omega t)\exp(-|t|) is an even function,

    I=\int_{-\infty}^{\infty}t\sin(\omega t)\exp(-|t|)\,\mathrm{d}t=2\int_0^{\infty}t\sin(\omega t)\exp(-t)\,\mathrm{d}t

    Then, one can use the expression of the sine in terms of exponential functions : \sin(\omega t)=\frac{\exp(i\omega t)-\exp(-i\omega t)}{2i} which gives

    \begin{aligned} I&=\int_0^{\infty}t\left(\frac{\exp(i\omega t)-\exp(-i\omega t)}{2i}\right)\exp(-t)\,\mathrm{d}t\\ &=\int_0^{\infty}t\left(\frac{\exp((i\omega-1) t)-\exp(-(1+i\omega) t)}{2i}\right)\,\mathrm{d}t\end{aligned}

    Using integration by parts one can get rid off the t which is outside the exponential functions.
    You are making hard work of this, see the other thread

    RonL
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by bcvw85 View Post
    Hi i'm lookin for the integral of t*exp(-abs(t))sin(wt) dt with limits at +infinity and -infinity

    Anyone available to help?
    \int_{-\infty}^{\infty}te^{-|t|}\sin(wt)\,dt

    Write it as two separate integrals:

    \int_{-\infty}^{0}te^{t}\sin(wt)\,dt+\int_{0}^{\infty}te^  {-t}\sin(wt)\,dt

    Apply integration by parts:


    let u=t and dv=e^{t}\sin(wt)\,dt

    Therefore, \,du=\,dt and v=\frac{1}{w^2+1}\cdot\left(e^t\sin(wt)-e^t\cos(wt)\right) (you get v by using integration by parts...again...


    Plugging into the integration by parts formula:

    \left.\left[\frac{te^t}{w^2+1}\cdot\left(\sin(wt)-\cos(wt)\right)\right]\right|_{-\infty}^{0}-\frac{1}{w^2+1}\int_{-\infty}^{0}\left(e^t\sin(wt)-e^t\cos(wt)\right)\,dt

    Before we move on, note that \left.\left[\frac{te^t}{w^2+1}\cdot\left(\sin(wt)-\cos(wt)\right)\right]\right|_{-\infty}^{0} converges to 0.

    Thus, we have left:

    -\frac{1}{w^2+1}\int_{-\infty}^{0}\left(e^t\sin(wt)-e^t\cos(wt)\right)\,dt

    Apply integration by parts (again) for these two integals:

    -\frac{1}{w^2+1}\int_{-\infty}^{0}e^t\sin(wt)+\frac{1}{w^2+1}\int_{-\infty}^{0}e^t\cos(wt)\,dt

    I will leave steps out, hoping that you can do it on your own.

    Now, the integrals should become \left.\left[\frac{2we^t\cos(wt)}{(w^2+1)^2}\right]\right|_{-\infty}^{0}-\left.\left[\frac{(w^2-1)e^t\sin(wt)}{(w^2+1)^2}\right]\right|_{-\infty}^{0}

    \left[\frac{2w}{(w^2+1)^2}-0\right]-\left[0-0\right]=\color{red}\boxed{\frac{2w}{(w^2+1)^2}}

    Since we only evaluated half of the integral, double it to get the answer:

    \therefore \int_{-\infty}^{\infty}te^{-|t|}\sin(wt)\,dt=\color{red}\boxed{\frac{4w}{(w^2+  1)^2}}

    Hope that this made sense!!
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