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Math Help - MHF Integral Bee!

  1. #76
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    Quote Originally Posted by Random Variable View Post
    It's not my turn, but no new integral has been posted. So I'm going to throw one out there.

     \int^{\infty}_{0} \frac{e^{-ax}-e^{-bx}}{x} \ dx \ \ (a, b >0)
    Easy one. Again, the magic differentiation, with respect to a or b !!! I would let others to finish.
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  2. #77
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    Quote Originally Posted by Random Variable View Post
    It's not my turn, but no new integral has been posted. So I'm going to throw one out there.

     \int^{\infty}_{0} \frac{e^{-ax}-e^{-bx}}{x} \ dx \ \ (a, b >0)
    we may use the result: fis continuous and f(+\infty)exists,then
    \int_0^{+\infty} \frac{f(ax)-f(bx)}{x} \ dx =(f(0)-f(+\infty))\ln \frac{b}{a}
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  3. #78
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    Quote Originally Posted by Krizalid View Post
    Try this one:

    Evaluate \lim_{n\to\infty}\int_0^{\frac\pi3}\frac{\sin^nx}{  \sin^nx+\cos^nx}\,dx.
     \frac{\pi}{6} is it right ?

    We can obtain  \int_a^{\frac{\pi}{2} - a } \frac{ dx}{ 1 + \cot^n(x) }  for all value of n

    and at the same time when  0 \leq x \leq \frac{\pi}{4} ,  \frac{1}{1 + \cot^n(x)} tends to 0 when n tends to a large value


    But i am not sure , if i sketch the graph , It is just like a step function ..
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  4. #79
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    No, but you're close.
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  5. #80
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    Quote Originally Posted by Krizalid View Post
    Try this one:

    Evaluate \lim_{n\to\infty}\int_0^{\frac\pi3}\frac{\sin^nx}{  \sin^nx+\cos^nx}\,dx.
    let  u = \frac{\pi}{ 2} - x

    then  du = -dx

     \cos(\pi/2 - u) = \sin u

     \sin(\pi/2 -u) = \cos u

     \int^{\pi/3}_{0}\frac{\sin^{n}(x)}{\sin^{n}x + \cos^{n}x} dx = \int^{\pi/2}_{\pi/6}\frac{\cos^{n}(u)}{\sin^{n}u + \cos^{n}u} du

    at which point I realize that this method doesn't work because of the upper limit not being  \pi/2
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  6. #81
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    Quote Originally Posted by Krizalid View Post
    No, but you're close.
    Oh , the integral  \lim_{n\to\infty} \int_0^{\frac{\pi}{3}} \frac{ \sin^n(x)}{ \sin^n(x) + \cos^n(x)}~dx and

      \int_0^{\frac{\pi}{3}} \lim_{n\to\infty}  \frac{ \sin^n(x)}{ \sin^n(x) + \cos^n(x)}~dx

    are different ?

    Or oh i have made a mistake , it should be  \frac{\pi}{12} ?
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  7. #82
    Super Member Random Variable's Avatar
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    Quote Originally Posted by ynj View Post
    we may use the result: fis continuous and f(+\infty)exists,then
    \int_0^{+\infty} \frac{f(ax)-f(bx)}{x} \ dx =(f(0)-f(+\infty))\ln \frac{b}{a}
    Yep, it's a Frullani integral.
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  8. #83
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    Quote Originally Posted by Random Variable View Post
    Yep, it's a Frullani integral.
    Proof of Frullani Integral (assuming f(x) is differentiable, which is stronger than continuous):

    <br />
I(a)=\int_0^{\infty} \frac{f(ax)-f(bx)}{x} \ dx<br />

    <br />
\frac{dI}{da}=\int_0^{\infty} f'(ax) dx = [f(\infty)-f(0)]\frac{1}{a}<br />

    <br />
I(a)=[f(\infty)-f(0)]\ln a + C\quad (1)<br />

    Obviously when a=b,\;  I=0, i.e. [f(\infty)-f(0)]\ln b+C=0\Rightarrow C=[f(0)-f(\infty)]\ln b\quad (2)

    Combine (1) and (2), we get the previous result.
    Last edited by luobo; August 22nd 2009 at 07:53 PM.
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  9. #84
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    Quote Originally Posted by simplependulum View Post
    Oh , the integral  \lim_{n\to\infty} \int_0^{\frac{\pi}{3}} \frac{ \sin^n(x)}{ \sin^n(x) + \cos^n(x)}~dx and

      \int_0^{\frac{\pi}{3}} \lim_{n\to\infty}  \frac{ \sin^n(x)}{ \sin^n(x) + \cos^n(x)}~dx

    are different ?

    Or oh i have made a mistake , it should be  \frac{\pi}{12} ?
    Such kind of integral, in general, after certain variable substitution (for example, t=\tan\frac{x}{2}), can be transformed into rational functions.
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  10. #85
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    Quote Originally Posted by Krizalid View Post
    Try this one:

    Evaluate \lim_{n\to\infty}\int_0^{\frac\pi3}\frac{\sin^nx}{  \sin^nx+\cos^nx}\,dx.
    as simplependulum mentioned (without a proof again! haha) the answer is \frac{\pi}{12}. more generally if \frac{\pi}{4}< \alpha < \frac{\pi}{2}, and I_n(\alpha)=\int_0^{\alpha}\frac{\sin^nx}{\sin^nx+  \cos^nx}\,dx, then \lim_{n\to\infty}I_n(\alpha)=\alpha - \frac{\pi}{4}.

    the reason is that I_n(\alpha)=\int_0^{\frac{\pi}{2}}\frac{\sin^nx}{\  sin^nx+\cos^nx}\,dx - \int_{\alpha}^{\frac{\pi}{2}}\frac{\sin^nx}{\sin^n  x+\cos^nx}\,dx=\frac{\pi}{4} - \int_{\alpha}^{\frac{\pi}{2}}\frac{1}{1+\cot^nx}\,  dx.

    now, since 0 \leq \cot x \leq \cot \alpha < 1 for \alpha \leq x \leq \frac{\pi}{2}, we have \lim_{n\to\infty}\int_{\alpha}^{\frac{\pi}{2}}\fra  c{1}{1+\cot^nx}\,dx=\int_{\alpha}^{\frac{\pi}{2}} dx = \frac{\pi}{2} - \alpha and therefore \lim_{n\to\infty} I_n(\alpha) = \alpha - \frac{\pi}{4}.

    -------------------------------------------------------------

    here's a problem: Solve for real values of x,y: \ \int_0^{\infty} \frac{\sin^2(xt) \sin^3(yt)}{t^3} \ dt = \frac{\pi x^2}{12}.
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  11. #86
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    here's a problem: Solve for real values of x,y: \ \int_0^{\infty} \frac{\sin^2(xt) \sin^3(yt)}{t^3} \ dt = \frac{\pi x^2}{12}.
    Message deleted
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  12. #87
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    Quote Originally Posted by NonCommAlg View Post
    the question is to find all (x,y) \in \mathbb{R}^2 for which the given equality holds. well, (0,0) is just one solution.
    I had thought the problem is a proof problem. It actually is equation solving. So I removed my post and you replied really fast. Are you here all day and every minute?
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  13. #88
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    It is time for you to apply " magic differentiation "
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  14. #89
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    and the last problem: Suppose a,b,c > 0. Evaluate \lim_{n\to\infty} \frac{1}{n^3} \int_1^e \ln(1 + ax^n) \ln(1 + bx^n) \ln(1 + cx^n) \ dx.
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  15. #90
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    Sorry , i forgot it , now let me post the solution to
     \int_0^{\pi} (\sin(x))^a \sin(ax)~dx

    consider

     (1/2i)^a \int_0^{\pi} e^{iax} ( e^{ix} - e^{-ix} )^a ~dx

     = (1/2i)^a \int_0^{\pi} ( e^{2ix} - 1 )^a ~dx

    Sub  2x = t

     = (1/2i)^a (1/2) \int_0^{2\pi} ( e^{it} - 1)^a ~ dt

    adn sub  z = e^{it}

    the integral becomes

     (1/2i)^{a} (1/2i) (-1)^a \int_{|z|=1} \frac{ (1-z)^a}{z} ~dz

     = (1/2)^a e^{ai(\pi - \frac{\pi}{2})} (1/2i) ( 2\pi i )

     = \frac{\pi}{2^a} e^{i \frac{\pi}{2} }

    Oh there is something wrong : the last line , it should be  = \frac{\pi}{2^a} e^{i \frac{a\pi}{2} }

    so   \int_0^{\pi} e^{iax} ( \sin(x))^a ~dx =  \frac{\pi}{2^a} e^{i \frac{\pi}{2} }

    compare the Im part  \int_0^{\pi} \sin(ax) ( \sin(x))^a ~dx =  \frac{\pi}{2^a} \sin(a\frac{\pi}{2})
    ( note : we can see  \sin(x) \geq 0     ,   0 \leq x \leq \pi then  (\sin(x))^a \in \mathbb{R^+} so that we can compare the real and imaginary part
    Last edited by simplependulum; August 23rd 2009 at 03:15 AM.
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