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Math Help - Integral

  1. #1
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    Integral

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    \int_{0}^{\pi} e^{r\cos x}\cos(x+r\sin x)\,dx=0
    Last edited by mr fantastic; September 18th 2009 at 09:04 AM. Reason: Restored original post
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  2. #2
    Super Member Matt Westwood's Avatar
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    I'd try substituting z = r \cos \theta and see where that got me. What have you tried so far?

    Sorry, ignore me - I didn't notice this was a challenge forum! Okay, I'll have a go ...
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  3. #3
    Super Member Matt Westwood's Avatar
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    Hang on, this is Bessel functions, isn't it? Way over my head, sorry about that. I'm going to have to give this one a miss till I've learned up on it.
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  4. #4
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    First, use the sum of angles formula on \cos(x+r\sin{x}):
    \cos(x+r\sin x)=\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x}),
    so our integral becomes:
    \int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\int_  0^{\pi}e^{r\cos{x}}\left[\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})\right]\,dx
    =\int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx
    Now, one can simply note that \frac{d}{dx}e^{r\cos{x}}=-e^{r\cos{x}}r\sin{x}, and that \frac{d}{dx}\sin(r\sin{x})=\cos(r\sin{x})r\cos{x}, so that our integral is
    \int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx =\int_0^{\pi}e^{r\cos{x}}\cdot\frac{1}{r}\left(\fr  ac{d}{dx}\sin(r\sin{x})\right)+\frac{1}{r}\left(\f  rac{d}{dx}e^{r\cos{x}}\right)\sin(r\sin{x})\,dx
    We recogize this as having the form u\frac{dv}{dx}+\frac{du}{dx}v=\frac{d}{dx}(uv), and so:
    \int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\frac  {1}{r}\int_0^{\pi}\frac{d}{dx}\left(e^{r\cos{x}}\s  in(r\sin{x})\right)\,dx
    =\frac{1}{r}\left[e^{r\cos{x}}\sin(r\sin{x})\right]_{x=0}^{\pi}
    =\frac{1}{r}\left(\frac{e^{r\cdot-1}\sin(r\cdot0)}{r}\right)-\frac{1}{r}\left(\frac{e^{r\cdot1}\sin(r\cdot0)}{r  }\right)=0

    -Kevin C.
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  5. #5
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    Quote Originally Posted by TwistedOne151 View Post
    First, use the sum of angles formula on \cos(x+r\sin{x}):
    \cos(x+r\sin x)=\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x}),
    so our integral becomes:
    \int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\int_  0^{\pi}e^{r\cos{x}}\left[\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})\right]\,dx
    =\int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx
    Now, one can simply note that \frac{d}{dx}e^{r\cos{x}}=-e^{r\cos{x}}r\sin{x}, and that \frac{d}{dx}\sin(r\sin{x})=\cos(r\sin{x})r\cos{x}, so that our integral is
    \int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx =\int_0^{\pi}e^{r\cos{x}}\cdot\frac{1}{r}\left(\fr  ac{d}{dx}\sin(r\sin{x})\right)+\frac{1}{r}\left(\f  rac{d}{dx}e^{r\cos{x}}\right)\sin(r\sin{x})\,dx
    We recogize this as having the form u\frac{dv}{dx}+\frac{du}{dx}v=\frac{d}{dx}(uv), and so:
    \int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\frac  {1}{r}\int_0^{\pi}\frac{d}{dx}\left(e^{r\cos{x}}\s  in(r\sin{x})\right)\,dx
    =\frac{1}{r}\left[e^{r\cos{x}}\sin(r\sin{x})\right]_{x=0}^{\pi}
    =\frac{1}{r}\left(\frac{e^{r\cdot-1}\sin(r\cdot0)}{r}\right)-\frac{1}{r}\left(\frac{e^{r\cdot1}\sin(r\cdot0)}{r  }\right)=0

    -Kevin C.
    nice! i like this integral better though: \int_0^{\pi} e^{r \cos x} \cos (r \sin x) \ dx.
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  6. #6
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    Quote Originally Posted by TwistedOne151 View Post
    First, use the sum of angles formula on \cos(x+r\sin{x}):
    \cos(x+r\sin x)=\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x}),
    so our integral becomes:
    \int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\int_ 0^{\pi}e^{r\cos{x}}\left[\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})\right]\,dx
    =\int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx
    Now, one can simply note that \frac{d}{dx}e^{r\cos{x}}=-e^{r\cos{x}}r\sin{x}, and that \frac{d}{dx}\sin(r\sin{x})=\cos(r\sin{x})r\cos{x}, so that our integral is
    \int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx =\int_0^{\pi}e^{r\cos{x}}\cdot\frac{1}{r}\left(\fr ac{d}{dx}\sin(r\sin{x})\right)+\frac{1}{r}\left(\f rac{d}{dx}e^{r\cos{x}}\right)\sin(r\sin{x})\,dx
    We recogize this as having the form u\frac{dv}{dx}+\frac{du}{dx}v=\frac{d}{dx}(uv), and so:
    \int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\frac {1}{r}\int_0^{\pi}\frac{d}{dx}\left(e^{r\cos{x}}\s in(r\sin{x})\right)\,dx
    =\frac{1}{r}\left[e^{r\cos{x}}\sin(r\sin{x})\right]_{x=0}^{\pi}
    =\frac{1}{r}\left(\frac{e^{r\cdot-1}\sin(r\cdot0)}{r}\right)-\frac{1}{r}\left(\frac{e^{r\cdot1}\sin(r\cdot0)}{r }\right)=0

    -Kevin C.
    That's right. I just happened to find out this interesting integral when I was doing the line integral posted here today by transgalactic, using both line integral and double integral.

    http://www.mathhelpforum.com/math-he...-question.html

    Now you have got this: \int_{0}^{a} e^{r\cos x}\cos(x+r\sin x)\,dx=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}, can you go ahead and find \int_{0}^{a} e^{r\cos x}\cos(2x+r\sin x)\,dx=?


    Last edited by mr fantastic; September 18th 2009 at 09:05 AM. Reason: Restored original reply
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  7. #7
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    Quote Originally Posted by luobo View Post

    Now you have got this: \int_{0}^{a} e^{r\cos x}\cos(x+r\sin x)\,dx=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}, can you go ahead and find \int_{0}^{a} e^{r\cos x}\cos(2x+r\sin x)\,dx=?
    sure! just differentiate both sides of \int_{0}^{a} e^{r\cos x}\cos(x+r\sin x)\,dx=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r} with respect to r.
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  8. #8
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    Quote Originally Posted by NonCommAlg View Post
    sure! just differentiate both sides of \int_{0}^{a} e^{r\cos x}\cos(x+r\sin x)\,dx=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r} with respect to r.
    To me, that's the beauty of this integral. You can continue the differentiation to find \int_{0}^{a} e^{r\cos x}\cos(nx+r\sin x)\,dx, n\geq 1 and you can also find the curve f(x)=e^{r\cos x}\cos(nx+r\sin x) is a beauty as well in [0, \pi] when you plot it out.

    \int_{0}^{a} e^{r\cos x}\cos(nx+r\sin x)\,dx=\frac{d^{n-1}}{dr^{n-1}}\left(\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}\righ  t), n\geq 1

    I am sure NonCommAlg has a solution for n=0, so can you please share with us?

    Additional question is: What happens if  n is zero or negative. Is there a closed-form solution. I guess not, if so, how about if a =\pi. I don't have a solution for this. If you have, can you please share.
    Last edited by mr fantastic; September 18th 2009 at 09:10 AM. Reason: Restored original reply and corrected some latex
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  9. #9
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    Quote Originally Posted by NonCommAlg View Post
    nice! i like this integral better though: \int_0^{\pi} e^{r \cos x} \cos (r \sin x) \ dx.

    Which is  \pi , independent of  r . I wonder if this one has a closed-form formula in terms of  r so that by taking derivative of it you get the case for n=1.
    Last edited by mr fantastic; September 18th 2009 at 09:07 AM. Reason: Restored original reply
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  10. #10
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    \int_{0}^{a} e^{r\cos x}\cos(nx+r\sin x)\,dx=\frac{d^{n-1}}{dr^{n-1}}\left(\frac{e^{r\cos{a}}\sin(r\sin{a})}{r}\righ  t), n\geq 1 (*)

    Let me give an example showing how the previous equality is used to get the summation of an important series.

    [/COLOR][/COLOR]
    f(r)=\frac{e^{r\cos{a}}\sin(r\sin{a})}{r} (1)

    According to the equality,
    f^{(n-1)}(r)=\int_{0}^{a} e^{r\cos x}\cos(nx+r\sin x)\,dx
    (2)

    Therefore,
    f^{(n-1)}(0)=\int_{0}^{a} \cos(nx)\,dx=\frac{\sin(na)}{n}
    (3)

    Taylor's series,
    f(r)=\sum_{n=1}^{\infty}\frac{f^{(n-1)}(0)}{(n-1)!}\,
    r^{n-1} (4)

    Plug (3) into (4),
    f(r)=\sum_{n=1}^{\infty}\frac{\sin(na)}{n!}\, r^{n-1} (5)

    Multiply r on both sides,
    rf(r)=\sum_{n=1}^{\infty}\frac{\sin(na)}{n!}\, r^{n} (6)

    Finally, the series:
    \sum_{n=1}^{\infty}\frac{\sin(na)}{n!}\,r^{n}=e^{r \cos a}\sin(r\sin a) (7)

    Some Special cases:
    (a) r=1
    \sum_{n=1}^{\infty}\frac{\sin(na)}{n!}=e^{\cos a}\sin(\sin a) (8)

    (b) r=1, a=1
    \sum_{n=1}^{\infty}\frac{\sin n}{n!}=e^{\cos 1}\sin(\sin1) (9)

    (c) a=\frac{\pi}{2}
    \sum_{n=1}^{\infty}\frac{\sin (n\pi/2)}{n!}r^n=\sin r (10) (which is within everybody's expectation)


    By Luobo
    08/14/2009
    Last edited by mr fantastic; September 18th 2009 at 09:13 AM. Reason: Restored original reply.
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  11. #11
    Senior Member pankaj's Avatar
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    Quote Originally Posted by TwistedOne151 View Post
    First, use the sum of angles formula on \cos(x+r\sin{x}):
    \cos(x+r\sin x)=\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x}),
    so our integral becomes:
    \int_0^{\pi}e^{r\cos{x}}\cos(x+r\sin{x})\,dx=\int_  0^{\pi}e^{r\cos{x}}\left[\cos{x}\cos(r\sin{x})-\sin{x}\sin(r\sin{x})\right]\,dx
    =\int_0^{\pi}e^{r\cos{x}}\cos{x}\cos(r\sin{x})-e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx


    -Kevin C.
    Consider \int_0^{\pi}e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx.

    Integrating by parts
    \int_0^{\pi}e^{r\cos{x}}\sin{x}\sin(r\sin{x})\,dx

    =\frac{1}{r}\left[-e^{r\cos x}\sin (r\sin x)\right]_{x=0}^{\pi}-\frac{1}{r}\left[\int_{0}^{\pi}(-e^{r\cos x}).\cos(r\sin x).r\cos x.dx\right]

    =\int_{0}^{\pi}e^{r\cos x}.\cos(r\sin x).\cos x.dx

    and you directly get 0
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