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Integral problems

  1. May 6, 2005 #1
    i think i don't have adequate theory to solve these problems.... :frown:

    1.given that f(x) =cos(x) sin^k(x) / (1+x). calculate integral of this function wrt x between limits 0 and pi/2 . then find the it's limit as k tends to infinity...

    2.let f(x)={e^(-ax)-e^(-bx)}/x, 0< a< b .let I be the integral of f wrt x between 0 and infinity. express I as a double integral...
     
  2. jcsd
  3. May 6, 2005 #2
    1) Are you meaning : [tex] \sin^k(x)\textrm{ or }\sin(x)^k [/tex] ?
     
  4. May 6, 2005 #3
    the former...
     
  5. May 6, 2005 #4

    dextercioby

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    Do you have to compute these 2

    [tex]F(k)=:\int_{0}^{\frac{\pi}{2}}\left[\cos x\left(\frac{\sin^{k}x}{1+x}\right)\right] dx [/tex]

    [tex] \lim_{k\rightarrow +\infty} F(k) [/tex]

    ...?

    Daniel.
     
  6. May 6, 2005 #5

    dextercioby

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    Here's the answer for

    [tex] F(1)=\frac {1}{2}\mbox{Si}\left( 2+\pi \right) \cos 2-\frac {1}{2}\mbox{Ci}\left( 2+\pi \right) \sin 2-\frac {1}{2}\mbox{Si}\left( 2\right) \cos 2+ \frac {1}{2}\mbox{Ci}\left( 2\right) \sin 2 [/tex]

    [tex] F(2)= -\frac{1}{4}\mbox{Si}\left( \frac {3}{2}\pi +3\right) \sin 3-\frac{1}{4}\mbox{Ci}\left( \frac {3}{2}\pi +3\right) \cos 3+\frac{1}{4}\mbox{Si}\left( 1+\frac{1}{2}\pi \right) \sin 1[/tex]
    [tex]+\frac{1}{4}\mbox{Ci}\left( 1+\frac{1}{2}\pi \right) \cos 1+\frac{1}{4}\mbox{Si}\left( 3\right) \sin 3+\frac{1}{4}\mbox{Ci}\left( 3\right) \cos 3-\frac{1}{4}\mbox{Si}\left( 1\right) \sin 1-\frac{1}{4}\mbox{Ci}\left( 1\right) \cos 1 [/tex]

    Daniel.
     
    Last edited: May 6, 2005
  7. May 6, 2005 #6

    dextercioby

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  8. May 6, 2005 #7
    Well the first is really unsolvable I think...even the latter is hard....maybe a start like this :

    [tex] \frac{sin(x)^kcos(x)}{1+x}=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^k\frac{e^{ix}+e^{-ix}}{2}\frac{1}{1+x} [/tex]

    [tex] = \frac{1}{2^{k+1}i^k}\sum_{n=0}^k\left(\begin{array}{c} k\\n\end{array}\right)(-1)^{n-k}\frac{e^{i(2n-k+1)x}}{1+x}+\frac{e^{i(2n-k-1)x}}{1+x} [/tex]

    Remain to calculate things of the form [tex] \int_0^{\frac{\pi}{2}}\frac{e^{imx}}{1+x}=\int_0^{\frac{\pi}{2}}e^{imx}\sum_{p=0}^\infty(-1)^p x^p [/tex]


    ....well no sorry, i get lost and wrong, maybe some contour integral with residues in the complex plane....or a recursion relation over k
     
  9. May 6, 2005 #8

    dextercioby

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    I linked you to a website which uses WebMathematica.I used my own antique Maple to come up with this (see attached file).

    I can only assume that the limit is 0...

    Daniel.
     
    Last edited: Nov 22, 2006
  10. May 6, 2005 #9

    shmoe

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    1. Do you need to find the integral exactly? A bound will suffice if it's just the limit you're interested in, 1/(1+x) is bounded by 1 on [0..pi/2], cos and sin are non-negative there.
     
  11. May 6, 2005 #10

    Zurtex

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    I've messed about in mathematica and got that:

    [tex]\int_0^{\frac{\pi}{2}} \frac{\cos x \sin^{1000000000} x}{1 + x} dx = 0 \quad \text{(to 100 dp)}[/tex]

    Also if you look at the graph of the function as you increase k for either the whole function or just sink (x) it starts to look like the area does tend towards 0.

    Perhaps this is about looking at the properties of the function, it would seem that:

    [tex]\lim_{k \rightarrow \infty} \sin^k x = \left\{ \begin{array}{rcl}
    1 & \mbox{for} & x = \pi \left(2n + \frac{1}{2} \right) \quad n \in \mathbb{Z}\\
    \text{Undefined} & \mbox{for} & x = \pi \left(2n - \frac{1}{2} \right) \quad n \in \mathbb{Z} \\
    0 & \mbox{for} & \text{elsewhere}
    \end{array}\right.
    [/tex]

    Now using that to look at the original function it's fairly easy to derive that:

    [tex]\lim_{k \rightarrow \infty} \frac{\cos x \sin^{k} x}{1 + x} = 0 \quad \forall x \in \left[0,\frac{\pi}{2}\right][/tex]

    Suddenly looks a lot easier to integrate to me o:)
     
    Last edited: May 6, 2005
  12. May 6, 2005 #11
    thanks for the help...
    never mind the second one...i figured it out.
     
  13. May 6, 2005 #12
    Do you know why we always write sin(x)^k=sin^k(x) ???

    This seems to me a notation abuse. Let's take k positive integer, then

    sin(x)^k=sin(x)*sin(x)*....sin(x) k times

    sin^k(x)=sin(sin(...sin(x)))...)) k times iterated

    Or you have another notation for iteration n-times ?
     
  14. May 7, 2005 #13
    sin^k(x)=(sinx)^k =sinx *sinx*.......k times
    and to express the second one let f(x)=sinx
    then f^k(x)= (sin(sin(....sin(x)))....))
     
  15. May 7, 2005 #14

    Zurtex

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    Actually it's not to confuse [itex]\sin (x^k)[/itex] with [itex](\sin x)^k[/itex] as most people don't bother writing out the brackets.

    But yeah notations don't seem to be entirely generalized.
     
  16. May 7, 2005 #15
    klein, I've never seen the sin function self-iterated like you demonstrated,

    [tex] sin^k(x) = sin(x)^k [/tex]

    The one on the RHS isn't used because of this one,
    and they arent equalto [tex] sin(x^k) [/tex]
     
  17. May 7, 2005 #16

    dextercioby

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    Actually the argument of [itex] \sin x [/itex] is never paranthesized,unless it gets really complicated...There's no point in writing [itex] \sin\left(x\right) [/itex],but it is for [itex] \sin\left(x+y\right) [/itex].

    Daniel.
     
  18. May 7, 2005 #17

    Zurtex

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    But that's my point, can you see the difference between:

    [tex]\sin {x^k}[/tex]

    And:

    [tex]{\sin x}^k[/tex]

    Unless you look at the code?
     
  19. May 7, 2005 #18

    dextercioby

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    I didn't look.Did u use "\displaystyle" ...?

    Daniel.

    EDIT:No,you didn't. :tongue2: I'm not trying to impose my own beliefs,just to stress out that writing

    [tex] \sin x^{k}\neq \left(\sin x\right)^{k}\equiv \sin^{k} x [/tex]

    has a motivation.If someone else sees it and agrees on it,then i'm happy...
     
    Last edited: May 7, 2005
  20. May 7, 2005 #19
    But it does bother me that [tex]\sin^2{x}=(\sin{x})^2[/tex] but [tex]\sin^{-1}{x}\neq \frac{1}{\sin{x}}[/tex]! Why use the same notation for these two concepts? Why does the k in sin^k(x) represent a functional power if k=-1, but not for any other k? Bah!
     
  21. May 7, 2005 #20
    Wonder what [itex] sin^{-2}x [/itex] would translate as?
     
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