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Evaluate this definite integral

  1. Sep 23, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    $$ \displaystyle \int_0^{\infty} e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2} da $$

    2. Relevant equations

    3. The attempt at a solution

    Evaluating this using integration by parts will be a cumbersome process and I don't even think that would give me the answer. Substitutions aren't going to make this problem easier. What other methods should I apply?
     
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  3. Sep 23, 2014 #2

    Simon Bridge

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    simplify or change the form of the integrand ... i.e. look for an identity involving the trig functions.
    the euler formula perhaps ... what have you tried?

    [edit] Ugh - looking up the indefinite integrals ... it's nasty.
    See "trigonometric integrals"
    http://en.wikipedia.org/wiki/Trigonometric_integral

    Where does this come from?
     
    Last edited: Sep 23, 2014
  4. Sep 23, 2014 #3

    SteamKing

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    Is the variable of integration 'a' instead of 'x'?
     
  5. Sep 23, 2014 #4

    utkarshakash

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    Yes
     
  6. Sep 23, 2014 #5

    Curious3141

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    Differentiating under the integral sign might help here, I think.

    $$\displaystyle \frac{d}{dx}\int_0^{\infty} e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2} da = \int_0^{\infty} \frac{\partial}{\partial x} (e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2}) da = \int_0^{\infty}e^{-x}\cos ax da $$ which is non convergent.

    So the original integral also does not exist.

    I'm not 100% sure about this, though.
     
    Last edited: Sep 23, 2014
  7. Sep 23, 2014 #6

    haruspex

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    Then I don't understand the significance of the exponential term. Why is that not just a factor that can be taken outside the integral? Or is everything inside the integral supposed to be forming the exponent?
     
  8. Sep 23, 2014 #7

    Ray Vickson

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    I looked at your profile, and so cannot believe this is a homework problem you have been given in a course. For that reason, I am going to chance giving you a complete solution, a practice I normally avoid.

    Let's change your unfortunate and rather horrible notation. What you call 'x' I will call 'b', and what you call 'a' I will call 'w'. So, you want to perform the integration
    [tex] \int_0^{\infty} e^{-b} \frac{w \sin(b w) - \cos(bw)}{1+w^2} \, dw [/tex]
    First: the constant ##e^{-b}## plays no role, so omit it; you can just do the integral without it, then put it back again in the final answer.
    Second: the integrand is an even function of ##w##, so ##\int_0^{\infty} = \frac{1}{2} \int_{-\infty}^{\infty}##, a fact that proves convenient later.

    OK, so we want the integral
    [tex] \int_{-\infty}^{\infty} \frac{ w \sin(bw) - \cos(bw)}{1+w^2} \, dw [/tex]
    WLOG, assume b > 0. We have ##w \sin(bw) = w \text{Re}[-i e^{ibw}]## and ##-\cos(bw) = \text{Re}[-e^{ibw}]##, so the integrand is the real part of
    [tex] F = \frac{-(1+iw) e^{ibw}}{1+w^2} = \frac{-(1+iw) e^{ibw}}{(1+iw)(1-iw)} =-i \frac{e^{ibw} }{w+i}. [/tex]
    Thus, it is enough to compute
    [tex] J = \int_{-\infty}^{\infty} \frac{e^{ibw}}{w+i} \, dw [/tex]


    This is an improper integral, so let's replace it by
    [tex] K_r = \int_{-\infty}^{\infty} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw [/tex]
    and then take the limit as ## r \to 0+##.

    The integrand of ##K_r## has a simple pole in the complex w-plane at the point ##w = -i##. For ##b > 0## we may complete the contour in the upper half w-plane, to get
    [tex] K_r = \lim_{R \to \infty} \int_{-R}^{R} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw
    = \lim_{R \to \infty} \oint_{\Gamma} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw,[/tex]
    where ##\Gamma## is the contour that goes along the real w-axis from ##w = -R+i0## to ##w = +R+i0##, then goes counterclockwise from ##R + i0## to ##-R + i0## along the semicircle of radius ##R## in upper half of the complex ##w##-plane. But the contour integral = 0 because the integrand is analytic inside ##\Gamma##. Thus, ##K_r = \lim_{R \to \infty} 0 = 0##, and this holds for all ##r > 0##. Taking the limit we have
    [tex] \int_{-\infty}^{\infty} \frac{e^{ibw}}{w+i} \, dw = \lim_{r \to 0} K_r = \lim_{r \to 0} 0 = 0.[/tex]

    Thus, we obtain
    [tex] \int_{-\infty}^{\infty} \frac{ w \sin(bw) - \cos(bw)}{w^2+1} \, dw = 0,[/tex]
    provided that we regard it as the limit where ##r \to 0+##.
     
  9. Sep 23, 2014 #8

    utkarshakash

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    I appreciate your effort and patience for posting the complete solution. However, your method was too complex for me. I don't know anything about contours and related terms. The integral which I posted here is a part of another integral which forms the original problem and here's it.

    Calculate [itex]\displaystyle \int_0^{\infty} e^{-x} \dfrac{\sin ax}{x} dx [/itex]

    Using Leibnitz's theorem for differentiation, I get
    $$\displaystyle \int_0^{\infty} e^{-x} \cos ax dx $$
    For evaluating this integral, I used the property(which is provided as a hint to the solution in my textbook) that
    [itex]\displaystyle \int e^{ax} \cos bx dx = \dfrac{e^{ax}(a\cos bx + b \sin bx)}{a^2+b^2} [/itex]

    Substituting appropriate values of a and b in the given formula, I get the integral posted in the thread.
     
  10. Sep 23, 2014 #9

    vela

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    You need to plug the limits in for ##x##. That'll simplify the integrand to the point where it's easy to integrate with respect to ##a##.
     
  11. Sep 24, 2014 #10
    This reasoning is not correct. If you consider the integral ##\displaystyle \int_0^{\infty} \frac{\sin (ax)}{x}\,dx## and differentiate with respect to ##a##, you get a non convergent integral but the original integral does exist!

    BTW, @utkarshakash, how did you get to the integral in OP from the original integral?
     
  12. Sep 24, 2014 #11

    Curious3141

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    Thank you. Yes, that's why I stated I was unsure if it was correct in my first post, but I posted it in case someone else could use the idea to come up with a correct solution.
     
  13. Sep 24, 2014 #12

    haruspex

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    From there, can you not express the cos in complex exponentials and just integrate?
     
  14. Sep 25, 2014 #13

    utkarshakash

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  15. Sep 25, 2014 #14

    vela

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    Or integrate by parts. However, the OP doesn't even need to do that as the hint provided gives the result of the indefinite integration. The problem is that by not plugging in the limits for ##x##, one is left with a complicated integrand in the last integration needed to get to the final solution.
     
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