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Math Help - MHF Integral Bee!

  1. #91
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    Quote Originally Posted by NonCommAlg View Post
    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.
    HaHa , we can apply magic differentiation in this problem Again !
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  2. #92
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by simplependulum View Post
     \int_0^{\pi} (\sin(x))^a \sin(ax)~dx = \frac{\pi}{2^a} e^{i \frac{\pi}{2} }
    Note that e^{i \frac{\pi}{2} }=i, so you're saying
     \int_0^{\pi} (\sin(x))^a \sin(ax)~dx = \frac{\pi i}{2^a} which is a surprising result for a real integral!
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  3. #93
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    Quote Originally Posted by luobo View Post
    Let me have a try. . I may be wrong. If so, please criticize.

    <br />
\sin^2(xt) = \frac{1-\cos(2xt)}{2}<br />

    <br />
\sin^3(yt) = \frac{3\sin(yt)-\sin(3yt)}{4}<br />

    I(x)= \int_0^{\infty} \frac{\sin^2(xt) \sin^3(yt)}{t^3} \ dt=\frac{1}{8} \int_0^\infty \frac{[1-\cos(2xt)][3\sin(yt)-\sin(3yt)]}{t^3}\;dt<br />

    \frac{dI}{dx}=\frac{1}{4} \int_0^\infty \frac{\sin(2xt)[3\sin(yt)-\sin(3yt)]}{t^2}\;dt<br />

    \frac{d^2I}{dx^2}=\frac{1}{2} \int_0^\infty \frac{\cos(2xt)[3\sin(yt)-\sin(3yt)]}{t}\;dt\quad<br />
(Note: second-order!!!)

    \frac{d^2I}{dx^2}=<br />
\frac{3}{4} \int_0^\infty \frac{\sin[(y+2x)t]+\sin[(y-2x)t]}{t}\;dt-<br />
<br />
\frac{1}{4} \int_0^\infty \frac{\sin[(3y+2x)t]+\sin[(3y-2x)t]}{t}\;dt\quad<br />

    \frac{d^2I}{dx^2}=<br />
\frac{3}{4} \int_0^\infty \frac{\sin[(y+2x)t]}{t}\;dt+<br />
\frac{3}{4} \int_0^\infty \frac{\sin[(y-2x)t]}{t}\;dt-<br />
<br />
\frac{1}{4} \int_0^\infty \frac{\sin[(3y+2x)t]}{t}\;dt-<br />
\frac{1}{4} \int_0^\infty \frac{\sin[(3y-2x)t]}{t}\;dt<br />

    \frac{d^2I}{dx^2}=<br />
\frac{3\pi}{8} [sign(y+2x)+sign(y-2x)] - \frac{\pi}{8} [sign(3y+2x)+sign(3y-2x)] <br />

    Since the original integration is even of x, so can assume x\geq0 (i.e, if x is a solution, so is -x). If so, there are totally five cases in the right half plane:
    (1)  -\infty < y \leq -2x,  \quad \frac{d^2I}{dx^2}=-\frac{\pi}{2}
    (2)      -2x < y \leq -\frac{2}{3}x, \quad \frac{d^2I}{dx^2}=\frac{\pi}{4}
    (3)  -\frac{2}{3}x < y \leq \frac{2}{3}x, \quad \frac{d^2I}{dx^2}=0
    (4)  \frac{2}{3}x <y \leq 2x, \quad \frac{d^2I}{dx^2}=-\frac{\pi}{4}
    (5)  2x < y< +\infty, \quad \frac{d^2I}{dx^2}=\frac{\pi}{2}

    The last equation tells me under a certain zone, the second-order derivative \frac{d^2I}{dx^2} is a constant, although not continuous.

    For simplicity, assume \frac{d^2I}{dx^2}=a, then \frac{dI}{dx}=ax+b and  I=\frac{1}{2}ax^2+bx+c

    Obviously
    b=0 since \frac{dI}{dx}=0\;@\;x=0
    c=0 since I=0\;@\;x=0.

    Therefore,
    <br />
I=\frac{1}{2}ax^2=\frac{\pi x^2}{12}<br />

    So x=0 seems to be the only solution.
    Sorry, this will not go through.
    Last edited by luobo; August 23rd 2009 at 02:46 PM.
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  4. #94
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    luobo : i am looking forward to seeing your solution , actually , I have been shocked to see how you (or someone ) applied magic differentiation in the Integral Bee !! It is really a powerful tool

    i think the solution is two straight lines passing throguh the origin (at,bt) and (-at,bt)
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  5. #95
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    Quote Originally Posted by NonCommAlg View Post
    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.
    Simply note that: <br />
\log \left( {1 + a \cdot x^n } \right) - \log \left( {a \cdot x^n } \right) = \log \left( {1 + \tfrac{1}<br />
{{a \cdot x^n }}} \right) = O\left( {\tfrac{1}<br />
{{a \cdot x^n }}} \right)<br />

    Hence: <br />
\log \left( {1 + a \cdot x^n } \right) = n \cdot \log \left( x \right) + O\left( 1 \right)<br />
as n tends to infinity. - x in [1, e] -

    And so: <br />
\log \left( {1 + a \cdot x^n } \right)\log \left( {1 + b \cdot x^n } \right)\log \left( {1 + c \cdot x^n } \right) = n^3  \cdot \log ^3 \left( x \right) + O\left( {n^2 } \right)<br />
- just multiply out-

    Thus: <br />
\int_1^e {\log \left( {1 + a \cdot x^n } \right)\log \left( {1 + b \cdot x^n } \right)\log \left( {1 + c \cdot x^n } \right)dx}  = n^3  \cdot \int_1^e {\log ^3 \left( x \right)dx}  + O\left( {n^2 } \right)<br />

    And so the answer turns out to be: \int_1^e {\log ^3 \left( x \right)dx} = 6 - 2e -by parts-
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  6. #96
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    Here's another solution to my proposed problem using a powered tool:

    \int_{0}^{\frac{\pi }{3}}{\frac{\sin ^{n}x}{\sin ^{n}x+\cos ^{n}x}\,dx}=\int_{0}^{\frac{\pi }{4}}{\frac{dx}{1+\cot ^{n}x}}+\int_{\frac{\pi }{4}}^{\frac{\pi }{3}}{\frac{dx}{1+\cot ^{n}x}}.

    Now since \frac1{1+\cot^nx}\le1 and g(x)=1 is an integrable function on x\in \left[ 0,\frac{\pi }{3} \right], besides on \left( 0,\frac{\pi }{4} \right] we get \frac1{1+\cot^nx}\to0 and on \left[ \frac{\pi }{4},\frac{\pi }{3} \right] we have \frac1{1+\cot^nx}\to1, hence the Dominated Convergence Theorem applies and \underset{n\to \infty}{\mathop{\lim }}\,\int_{0}^{\frac{\pi }{3}}{\frac{\sin ^{n}x}{\sin ^{n}x+\cos ^{n}x}\,dx}=\int_{0}^{\frac{\pi }{4}}{0\,dx}+\int_{\frac{\pi }{4}}^{\frac{\pi }{3}}{1\,dx}=\frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}.
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  7. #97
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    Quote Originally Posted by Krizalid View Post
    Here's another solution to my proposed problem using a powered tool:

    \int_{0}^{\frac{\pi }{3}}{\frac{\sin ^{n}x}{\sin ^{n}x+\cos ^{n}x}\,dx}=\int_{0}^{\frac{\pi }{4}}{\frac{dx}{1+\cot ^{n}x}}+\int_{\frac{\pi }{4}}^{\frac{\pi }{3}}{\frac{dx}{1+\cot ^{n}x}}.

    Now since \frac1{1+\cot^nx}\le1 and g(x)=1 is an integrable function on x\in \left[ 0,\frac{\pi }{3} \right], besides on \left( 0,\frac{\pi }{4} \right] we get \frac1{1+\cot^nx}\to0 and on \left[ \frac{\pi }{4},\frac{\pi }{3} \right] we have \frac1{1+\cot^nx}\to1, hence the Dominated Convergence Theorem applies and \underset{n\to \infty}{\mathop{\lim }}\,\int_{0}^{\frac{\pi }{3}}{\frac{\sin ^{n}x}{\sin ^{n}x+\cos ^{n}x}\,dx}=\int_{0}^{\frac{\pi }{4}}{0\,dx}+\int_{\frac{\pi }{4}}^{\frac{\pi }{3}}{1\,dx}=\frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}.
    Actually, we can use the equation
    \lim_{n\rightarrow \infty}\int_{0}^{\frac{\pi }{3}}{\frac{\sin ^{n}x}{\sin ^{n}x+\cos ^{n}x}\,dx}=\int_{0}^{\frac{\pi }{3}}{\lim_{n\rightarrow \infty}\frac{\sin ^{n}x}{\sin ^{n}x+\cos ^{n}x}\,dx}without any doubt.
    A theorem says that if f_{n}(x)uniformly converges on [a,b], then
    \lim_{n\rightarrow \infty}\int_{a}^{b} {f_{n} (x) \,dx}=\int_{a}^{b} {\lim_{n\rightarrow \infty}f_{n} (x) \,dx}
    Now let \frac{\sin ^{n}x}{\sin ^{n}x+\cos ^{n}x}=f_n (x)
    In this problem, we can easily say that \forall\epsilon>0, f_n (x)uniformly converges on [0,\frac{\pi}{4}-\epsilon],[\frac{\pi}{4}+\epsilon,\frac{\pi}{3}],so \lim_{n\rightarrow \infty}\int_{0}^{\frac{\pi}{4}-\epsilon}{ f_{n}(x) \,dx}=\int_{0}^{\frac{\pi}{4}-\epsilon}{\lim_{n\rightarrow \infty} f_{n}(x) \,dx}=0, \lim_{n\rightarrow \infty}\int_{\frac{\pi}{4}+\epsilon}^{\frac{\pi}{3  }}{ f_{n}(x) \,dx}=\int_{\frac{\pi}{4}+\epsilon}^{\frac{\pi}{3}  }{\lim_{n\rightarrow \infty} f_{n}(x) \,dx}=\frac{\pi}{12}-\epsilon
    so \lim_{n\rightarrow \infty}\int_{0}^{\frac{\pi }{3}}f_n (x) \ dx=\frac{\pi}{12}-\epsilon+\lim_{n\rightarrow \infty}\int_{\frac{\pi}{4}-\epsilon}^{\frac{\pi }{4}+\epsilon}f_n (x) \ dx
    but |f_n (x)|<10\forall n,x,so |\lim_{n\rightarrow \infty}\int_{0}^{\frac{\pi }{3}}f_n (x) \ dx-\frac{\pi}{12}|\leq 19\epsilon
    let \epsilon\rightarrow 0, we can get \lim_{n\rightarrow \infty}\int_{0}^{\frac{\pi }{3}}f_n (x) \ dx=\frac{\pi}{12}!!
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  8. #98
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    Quote Originally Posted by NonCommAlg View Post

    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}.
    finding all x,y for which the above equality holds requires considering several cases. i'm not going to do that! haha so i will only evaluate the integral.

    let f(t)=\sin^2(xt) \sin^3(yt). then:

    1) \lim_{t\to 0} \frac{f(t)}{t^2}=\lim_{t\to\infty} \frac{f(t)}{t^2}=\lim_{t\to 0} \frac{f'(t)}{t}=\lim_{t\to\infty} \frac{f'(t)}{t}=0. therefore applying integration by parts twice gives us: \int_0^{\infty} \frac{f(t)}{t^3} \ dt =\frac{1}{2} \int_0^{\infty} \frac{f''(t)}{t} \ dt.

    2) we know that \int_0^{\infty} \frac{\sin(ct)}{t} \ dt = \frac{\pi}{2}\text{sgn}(c), where \text{sgn} is the sign function.

    3) using some routine trigonometry identities we get: 16f(t)=6\sin(yt) - 2 \sin(3yt) + \sin(3y+2x)t +  \sin(3y-2x)t -3\sin(y+2x)t -3 \sin(y-2x)t.

    4) applying 1) to 3) and then using 2) will finally give us:

    \int_0^{\infty} \frac{f(t)}{t^3} \ dt = \frac{\pi}{64}[ 12y^2 \text{sgn}(y) \ - \ (3y+2x)^2 \text{sgn} (3y+2x) \ - \ (3y-2x)^2 \text{sgn}(3y-2x) \ +

    3(y+2x)^2 \text{sgn}(y+2x)+3(y-2x)^2\text{sgn}(y-2x) ]. \ \Box
    Last edited by NonCommAlg; August 24th 2009 at 01:40 AM.
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  9. #99
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    Quote Originally Posted by NonCommAlg View Post
    nope! that's not the correct answer! the solution set is the line y=\frac{2}{3}x.
    But if (x,y) is the solution , then ( -x,y) is also the solution
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  10. #100
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    Quote Originally Posted by NonCommAlg View Post
    Solve for real values of
    Quote Originally Posted by NonCommAlg View Post
    nope! that's not the correct answer! the solution set is the line y=\frac{2}{3}x.

    The problem would be easier to solve if it changes this way:

    Solve for real values of
    <br />
x,u: \int_0^{\infty} \frac{\sin^2(xt) \sin^3(uxt)}{t^3} \ dt = \frac{\pi x^2}{12}

    or even

    Step (1): Show for a fixed u, the following integral is proportional to  x^2 , i.e.

     \int_0^{\infty} \frac{\sin^2(xt) \sin^3(uxt)}{t^3} \ dt = f(u)\;x^2<br />

    Step (2): Solve for real values of  u such that

    f(u) = \frac{\pi}{12}

    For Step (1), can we use the "differentiation"? hahaha...
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  11. #101
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    Quote Originally Posted by simplependulum View Post

    But if (x,y) is the solution , then ( -x,y) is also the solution
    yes, you're right about this but your answer is still wrong! see my previous post, which i just edited!

    to moderators: feel free to close this thread whenever you like! cheers to everybody!
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  12. #102
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    Quote Originally Posted by simplependulum View Post
    luobo : i am looking forward to seeing your solution , actually , I have been shocked to see how you (or someone ) applied magic differentiation in the Integral Bee !! It is really a powerful tool

    i think the solution is two straight lines passing throguh the origin (at,bt) and (-at,bt)
    simplependulum, the thread starter has put his solution, now it is time to put yours here, of course, not using differentiation! .
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  13. #103
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    Quote Originally Posted by NonCommAlg View Post
    finding all x,y for which the above equality holds requires considering several cases. i'm not going to do that! haha so i will only evaluate the integral.

    let f(t)=\sin^2(xt) \sin^3(yt). then:

    1) \lim_{t\to 0} \frac{f(t)}{t^2}=\lim_{t\to\infty} \frac{f(t)}{t^2}=\lim_{t\to 0} \frac{f'(t)}{t}=\lim_{t\to\infty} \frac{f'(t)}{t}=0. therefore applying integration by parts twice gives us: \int_0^{\infty} \frac{f(t)}{t^3} \ dt =\frac{1}{2} \int_0^{\infty} \frac{f''(t)}{t} \ dt.

    2) we know that \int_0^{\infty} \frac{\sin(ct)}{t} \ dt = \frac{\pi}{2}\text{sgn}(c), where \text{sgn} is the sign function.

    3) using some routine trigonometry identities we get: 16f(t)=6\sin(yt) - 2 \sin(3yt) + \sin(3y+2x)t +  \sin(3y-2x)t -3\sin(y+2x)t -3 \sin(y-2x)t.

    4) applying 1) to 3) and then using 2) will finally give us:

    \int_0^{\infty} \frac{f(t)}{t^3} \ dt = \frac{\pi}{64}[ 12y^2 \text{sgn}(y) \ - \ (3y+2x)^2 \text{sgn} (3y+2x) \ - \ (3y-2x)^2 \text{sgn}(3y-2x) \ +

    3(y+2x)^2 \text{sgn}(y+2x)+3(y-2x)^2\text{sgn}(y-2x) ]. \ \Box
    I was just overwhelmed by those different cases.
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  14. #104
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    Quote Originally Posted by NonCommAlg View Post
    yes, you're right about this but your answer is still wrong! see my previous post, which i just edited!

    to moderators: feel free to close this thread whenever you like! cheers to everybody!
    Nice thread. Thanks NonCommAlg.

    Closed.
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