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Thread: Per partes integral problem

  1. #1
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    Per partes integral problem

    I am calculating this integral:

    I have used the method per partes, but the problem is that I got something else. Is the math correct in the first red rectangle?



    How do I get that in the second red rectangle?
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  2. #2
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    Re: Per partes integral problem

    $\int t e^{\alpha t}~dt \neq \dfrac{t e^{\alpha t}}{\alpha}$

    $\int t e^{\alpha t}~dt =\dfrac{e^{\alpha t} (\alpha t-1)}{\alpha ^2}$
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  3. #3
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    Re: Per partes integral problem

    The first rectangle looks OK to me, but I'm less convinced about the next line.
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  4. #4
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    Re: Per partes integral problem

    Quote Originally Posted by romsek View Post
    $\int t e^{\alpha t}~dt \neq \dfrac{t e^{\alpha t}}{\alpha}$

    $\int t e^{\alpha t}~dt =\dfrac{e^{\alpha t} (\alpha t-1)}{\alpha ^2}$
    Well by this I also got with per partes method:

    $\frac{te^{j n \pi t}}{j n \pi} - \frac{e^{j n \pi t}}{j n \pi}$

    But the first term here, it's not in integral anymore, so we don't insert the interval numbers. First term is consequence of $f(x)g(x)$ from integrations by parts.
    So where does $t$ in the first term disapear if we don't insert interval numbers?

    Thank you.
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  5. #5
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    Re: Per partes integral problem

    Sorry, is my question here not clear?


    $\frac{te^{j n \pi t}}{j n \pi} = f(x)g(x)$

    This term remains in the equation. So where does "$t$" go in the calculation from my exercices?
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  6. #6
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    Re: Per partes integral problem

    $\begin{align*}
    &\dfrac 1 2 \displaystyle \int_0^1 ~(t-1)e^{-j n \pi t}~dt = \\ \\

    &\dfrac 1 2 \displaystyle \int_0^1 ~t e^{-j n \pi t}~dt - \displaystyle \int_0^1 ~e^{-j n \pi t}~dt

    \end{align*}$

    For the left hand term using per partes

    $u = t,~du = dt$

    $dv = e^{-j n \pi t},~v = -\dfrac{e^{-j n \pi t}}{j n \pi}$

    $\begin{align*}

    &\dfrac 1 2 \displaystyle \int_0^1 ~t e^{-j n \pi t}~dt = \\ \\

    &\left . -t \dfrac{e^{-j n \pi t}}{j n \pi} \right|_0^1 + \displaystyle \int_0^1 ~\dfrac{e^{-j n \pi t}}{j n \pi}~dt = \\ \\

    &\dfrac{j e^{-j n \pi}}{n\pi} - \left . \dfrac{e^{-j n \pi t}}{(j n \pi)^2} \right |_0^1 = \\ \\

    &\dfrac{j e^{-j n \pi}}{n\pi} + \left( \dfrac{e^{-j n \pi}-1}{n^2 \pi^2} \right)

    \end{align*}$

    The right hand term is just

    $\displaystyle \int_0^1~-e^{-j n \pi t}~dt = \dfrac{j(1-e^{-j n \pi})}{n \pi}$

    Combining these we get

    $\begin{align*}
    &\dfrac 1 2 \displaystyle \int_0^1 ~(t-1)e^{-j n \pi t}~dt = \\ \\

    &\dfrac{j}{n\pi}+ \left( \dfrac{e^{-j n \pi}-1}{n^2 \pi^2} \right) =\\ \\

    &\dfrac{e^{-j n \pi}-1 + j n \pi}{n^2\pi^2}

    \end{align*}$

    you can mess with the rest of the simplification

    $e^{j n \pi} = \begin{cases} 1 &n \text{ even} \\ -1 &n \text{ odd} \end{cases}$
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  7. #7
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    Re: Per partes integral problem

    Thanks romsek, for your effort and time.

    Ok, it took me a little more time.

    The thing is, I to this moment didn't understand the rule integration by parts, but for the definite integral, not indefinite. Let's took a closer look to the rule:

    $\int f(x)g(x)' = f(x)g(x) - \int f(x)' g(x)$

    I thought and still think that the $f(x)g(x)$ term does not need intervals to insert.
    What about the second red rectangle in my question?

    Does any one know? Thanks.
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  8. #8
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    Re: Per partes integral problem

    $A_n = \dfrac{(-1)^n - 1 + j n \pi}{2n^2 \pi^2}$

    $A_{-1} = \dfrac{-2 - j \pi}{2\pi^2}$

    $A_{1} = \dfrac{-2 + j \pi}{2\pi^2}$

    $\left|A_{-1} \right|^2 = \dfrac{4+\pi^2}{4\pi^4}$

    $\left|A_{1} \right|^2 = \dfrac{4+\pi^2}{4\pi^4}$

    $\left|A_{-1} \right|^2 = \left|A_{1} \right|^2$

    $\left|A_{-1} \right|^2 + \left|A_{1} \right|^2 = 2 \left|A_{1} \right|^2 = 2\dfrac{4+\pi^2}{4\pi^4} = \dfrac{4+\pi^2}{2\pi^4}$
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  9. #9
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    Re: Per partes integral problem

    Quote Originally Posted by Nforce View Post
    Thanks romsek, for your effort and time.

    Ok, it took me a little more time.

    The thing is, I to this moment didn't understand the rule integration by parts, but for the definite integral, not indefinite. Let's took a closer look to the rule:

    $\int f(x)g(x)' = f(x)g(x) - \int f(x)' g(x)$

    I thought and still think that the $f(x)g(x)$ term does not need intervals to insert.
    What about the second red rectangle in my question?

    Does any one know? Thanks.
    $\displaystyle \int_a^b f(x)g'(x)dx = f(b)g(b)-f(a)g(a) - \int_a^b f'(x)g(x)dx$
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  10. #10
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    Re: Per partes integral problem

    Quote Originally Posted by romsek View Post
    $A_n = \dfrac{(-1)^n - 1 + j n \pi}{2n^2 \pi^2}$

    $A_{-1} = \dfrac{-2 - j \pi}{2\pi^2}$

    $A_{1} = \dfrac{-2 + j \pi}{2\pi^2}$

    $\left|A_{-1} \right|^2 = \dfrac{4+\pi^2}{4\pi^4}$

    $\left|A_{1} \right|^2 = \dfrac{4+\pi^2}{4\pi^4}$

    $\left|A_{-1} \right|^2 = \left|A_{1} \right|^2$

    $\left|A_{-1} \right|^2 + \left|A_{1} \right|^2 = 2 \left|A_{1} \right|^2 = 2\dfrac{4+\pi^2}{4\pi^4} = \dfrac{4+\pi^2}{2\pi^4}$
    Ok, so we just insert for "n", but what does these two -1 and 1 represent from engineering point of view?
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  11. #11
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    Re: Per partes integral problem

    Quote Originally Posted by Nforce View Post
    Ok, so we just insert for "n", but what does these two -1 and 1 represent from engineering point of view?
    these are the coefficients that apply at frequencies $\pm 1 \cdot \nu$

    In this case $\nu = \dfrac 1 2$
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  12. #12
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    Re: Per partes integral problem

    Ok, this was a short answer, so more coeficients the better the aproximation of the signal. But if we just put aside the aproximation, for what else is this useful?
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  13. #13
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    Re: Per partes integral problem

    Quote Originally Posted by Nforce View Post
    Ok, this was a short answer, so more coeficients the better the aproximation of the signal. But if we just put aside the aproximation, for what else is this useful?
    well....

    in the physical world these sinusoids correspond to real vibrations.

    Suppose you had a piece of machinery that vibrated with a given signal, and further that lower frequency vibrations have the potential to cause damage.

    You'd like to find a way to damp those frequencies so your machine doesn't shake itself to death.

    But it's possible that some vibrations are significantly stronger at a couple of frequencies so it's much more cost efficient to just damp those few frequencies. Such damping corresponds to actual products that cost actual money.

    So to do this you need to know what frequencies you really want to damp and thus frequency analysis is useful.

    You've probably seen a spectrogram or possible a frequency analyzer before. Well Fourier analysis is how it gets those levels for each frequency.

    This is how an audio equalizer works for example, by allowing you to damp off or amplify your choice of frequencies.

    There are whole libraries and factories full of other examples.
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  14. #14
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    Re: Per partes integral problem

    It's also important in the construction of bridges and buildings for the same reason (except that the forcing signal is wind or earthquake action).
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  15. #15
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    Re: Per partes integral problem

    Well, ok then. Thanks.
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