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  1. #31
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by red_dog View Post
    1) Let f,g:[-a,a]\to\mathbb{R} two continuous functions, f odd, g even. Then

    \int_{-a}^a\frac{g(x)}{1+f(x)+\sqrt{1+f^2(x)}}dx=\int_{-a}^af(x)dx

    2) Calculate \int_{-1}^1\frac{x^{2008}}{1+x^{2007}+\sqrt{1+x^{4014}}}d  x
    can I use question one to solve question two without proving one ??

    or I should prove one then I can use it
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  2. #32
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    Quote Originally Posted by Amer View Post
    can I use question one to solve question two without proving one ??

    or I should prove one then I can use it
    i don't think question 2) needs any solution, since it's just a trivial application of 1).
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  3. #33
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by PaulRS View Post
    It should be <br />
\int_{ - a}^a {\tfrac{{g\left( x \right)}}<br />
{{1 + f\left( x \right) + \sqrt {1 + f^2 \left( x \right)} }}dx}  = \int_0^a {g\left( x \right) \cdot dx} <br />
    Yes, you're right. Sorry. I've edited.
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  4. #34
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    Quote Originally Posted by red_dog View Post
    1) Let f,g:[-a,a]\to\mathbb{R} two continuous functions, f odd, g even. Then

    \int_{-a}^a\frac{g(x)}{1+f(x)+\sqrt{1+f^2(x)}}dx=\int_0^a  g(x)dx
    ok, let h(x)=\frac{g(x)}{1+f(x)+\sqrt{1+f^2(x)}}. see that h(x) + h(-x) = g(x). also substituting x \to -x gives us \int_{-a}^0 h(x) \ dx = \int_0^a h(-x) \ dx. thus:

    \int_{-a}^a h(x) \ dx = \int_{-a}^0 h(x) \ dx + \int_0^a h(x) \ dx = \int_0^a (h(x) + h(-x)) \ dx = \int_0^a g(x) \ dx.



    Problem: Let f: [0, \infty) \longrightarrow \mathbb{R} be continuous and suppose that \lim_{x\to\infty} \left(f(x) + \int_0^x f(t) \ dt \right) exists and it's finite. Prove that \lim_{x\to\infty} f(x) = 0.
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  5. #35
    ynj
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    Let g(x)=\int_0^x f(t) \ dt
    Then \lim g(x)=\lim \frac{e^x g(x)}{e^x}=\lim\frac{e^x g(x)+e^x g'(x)}{e^x}=\lim g(x)+g'(x)=b,so \lim f(x)=\lim g'(x)=0
    My problem:
    Let fbe a polynomial of degree n. \int_0^1 f(x)x^k \ dx=0, k=1,2,...n
    Prove: \int_0^1 f^2(x) \ dx=(n+1)^2(\int_0^1 f(x) \ dx)^2
    Last edited by ynj; August 19th 2009 at 07:16 PM.
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  6. #36
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    Let  f(x) be  \sum_{k=0}^n a_k x^k

     \int_0^1 f^2(x)~dx = \int_0^1 f(x)[ \sum_{k=0}^n a_k x^k]~dx

     = a_0 \int_0^1 f(x)~dx

    And since  \int_0^1 \sum_{k=0}^n a_k x^{k+r}~dx = \sum_{k=0}^n \frac{a_k}{x+r+1} = 0 ~~ r=1,2,3...,n

     \int_0^1 f(x) x^r ~dx can be written as

     \frac{k(r-1)(r-2)(r-3)...(r-n)}{ (r+1)(r+2)...(r+n+1)}   ~~~~~(1)



    but the numerator is also equal to  \sum_{k=0}^n a_k [\prod_{j\neq k+1}^n (r+j)]~~~(2)

    Then sub  r = 0 and  r= -1 into  (1) and  (2) respectively , obtaining

     \int_0^1 f(x)~dx = \frac{ k (-1)^n}{n+1}

    and

     a_0 (n!) = k (-1)^n (n+1)! = (n+1)! \int_0^1 f(x)~dx (n+1)

    therefore

     \int_0^1 f^2(x)~dx = a_0 \int_0^1 f(x)~dx

     = (n+1)^2 ( \int_0^1 f(x)~dx)^2



    Problem :

     L^{-1} [ \sqrt{s + \sqrt{s^2 + a^2}}  ]

    Is it related to INTEGRAL ? If you think it is not , i still have another problem
    Last edited by simplependulum; August 19th 2009 at 11:21 PM.
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  7. #37
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    Quote Originally Posted by ynj View Post
    Let g(x)=\int_0^x f(t) \ dt
    Then \lim g(x)=\lim \frac{e^x g(x)}{e^x}=\lim\frac{e^x g(x)+e^x g'(x)}{e^x}=\lim g(x)+g'(x)=b,so \lim f(x)=\lim g'(x)=0
    good thinking but the problem is not that simple! your proof is not correct because you cannot use L'Hopital unless you know that \lim_{x\to\infty}e^x g(x) = \infty.
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  8. #38
    ynj
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    Quote Originally Posted by NonCommAlg View Post
    good thinking but the problem is not that simple! your proof is not correct because you cannot use L'Hopital unless you know that \lim_{x\to\infty}e^x g(x) = \infty.
    not exactly, L'Hospital do not require e^xg(x)->inf,buf just e^x->inf
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  9. #39
    ynj
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    maybe you will be confused about what i have said,here is a proof
    let \lim\frac{f'(x)}{g'(x)}=l(it exists)
    and the only restriction is that \lim g(x)=\infty,g(x)!=0
    now we will show \lim\frac{f(x)}{g(x)}=l
    \forall\epsilon>0,\exists\delta>0,\forall x>\delta,l-\epsilon<\frac{f'(x)}{g'(x)}<l+\epsilon
    so, for some (c,x)\subset (\delta,\infty), \exists\xi\in (c,x)
    l-\epsilon<\frac{f(x)-f(c)}{g(x)-g(c)}=\frac{f'(\xi)}{g'(\xi)}<l+\epsilon
    but \frac{f(x)-f(c)}{g(x)-g(c)}=(\frac{f(x)}{g(x)}-\frac{f(c)}{g(x)})(1-\frac{g(c)}{g(x)})^{-1}
    thus \frac{f(x)}{g(x)}<\frac{f(c)}{g(x)}+(l+\epsilon)(1-\frac{g(c)}{g(x)})
    let x->inf (sup inf), then
    \limsup\frac{f(x)}{g(x)}\leq l+\epsilon
    let epsilon->0, then
    \limsup\frac{f(x)}{g(x)}\leq l
    similarily,
    \liminf\frac{f(x)}{g(x)}\geq l
    so
    \lim\frac{f(x)}{g(x)}=l
    The proof is copyed from my book. Usually we require numerator to ->inf, but it is unnecessary.
    Last edited by ynj; August 19th 2009 at 10:56 PM.
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  10. #40
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    Quote Originally Posted by ynj View Post
    maybe you will be confused about what i have said,here is a proof
    let \lim\frac{f'(x)}{g'(x)}=l(it exists)
    and the only restriction is that \lim g(x)=\infty,g(x)!=0
    now we will show \lim\frac{f(x)}{g(x)}=l
    \forall\epsilon>0,\exists\delta>0,\forall x>\delta,l-\epsilon<\frac{f'(x)}{g'(x)}<l+\epsilon
    so, for some (c,x)\subset (\delta,\infty), \exists\xi\in (c,x)
    l-\epsilon<\frac{f(x)-f(c)}{g(x)-g(c)}=\frac{f'(\xi)}{g'(\xi)}<l+\epsilon
    but \frac{f(x)-f(c)}{g(x)-g(c)}=(\frac{f(x)}{g(x)}-\frac{f(c)}{g(x)})(1-\frac{g(c)}{g(x)})^{-1}
    thus \frac{f(x)}{g(x)}<\frac{f(c)}{g(x)}+(l+\epsilon)(1-\frac{g(c)}{g(x)})
    let x->inf (sup inf), then
    \limsup\frac{f(x)}{g(x)}\leq l+\epsilon
    let epsilon->0, then
    \limsup\frac{f(x)}{g(x)}\leq l
    similarily,
    \liminf\frac{f(x)}{g(x)}\geq l
    so
    \lim\frac{f(x)}{g(x)}=l
    The proof is copyed from my book. Usually we require numerator to ->inf, but it is unnecessary.
    i see! you used this theorem! you should've mentioned that in your solution. good work!
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  11. #41
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    Problem :

     L^{-1} [ \sqrt{s + \sqrt{s^2 + a^2}}  ]

    Is it related to INTEGRAL ? If you think it is not , i still have another problem
    OK. Let me have a try (first time to do an inverse LaPlace transform). Not sure if it is correct.

    Suppose  a>0, Re(s)>0

    <br />
F(s)\,=\, \sqrt{s + \sqrt{s^2 + a^2}} <br />

    <br />
F'(s)\,=\, \frac{\sqrt{s+\sqrt{s^2+a^2}}}{2\sqrt{s^2+a^2}}\,=  \, \frac{a^{\frac{1}{2}}}{2}\cdot\frac{a^{-\frac{1}{2}}(s+\sqrt{s^2+a^2})^{\frac{1}{2}}}{\sqr  t{s^2+a^2}}<br />

    <br />
L^{-1}[F'(s)]\,=\,\frac{a^{\frac{1}{2}}}{2}\cdot J_{-\frac{1}{2}}(at)\cdot u(t)<br />

    <br />
L^{-1}[F(s)]\,=-\,\frac{a^{\frac{1}{2}}}{2}\cdot J_{-\frac{1}{2}}(at)\cdot \frac{u(t)}{t}<br />
    Last edited by luobo; August 20th 2009 at 07:32 PM.
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  12. #42
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    Problem: Let f: [0, \infty) \longrightarrow \mathbb{R} be continuous and suppose that \lim_{x\to\infty} \left(f(x) + \int_0^x f(t) \ dt \right) exists and it's finite. Prove that \lim_{x\to\infty} f(x) = 0.[/quote]

    The solution is here.
    l'H˘pital's rule - Wikipedia, the free encyclopedia
    http://upload.wikimedia.org/math/5/f...c513fb6556.png

    Check last paragraph in Section "Examples".
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  13. #43
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    Quote Originally Posted by luobo View Post
    OK. Let me have a try (first time to do an inverse Laplace transform). Not sure if it is correct.

    Suppose  a>0, Re(s)>0

    <br />
F(s)\,=\, \sqrt{s + \sqrt{s^2 + a^2}} <br />

    <br />
F'(s)\,=\, \frac{\sqrt{s+\sqrt{s^2+a^2}}}{2\sqrt{s^2+a^2}}\,=  \, \frac{a^{\frac{1}{2}}}{2}\cdot\frac{a^{-\frac{1}{2}}(s+\sqrt{s^2+a^2})^{\frac{1}{2}}}{\sqr  t{s^2+a^2}}<br />

    <br />
L^{-1}[F'(s)]\,=\,\frac{a^{\frac{1}{2}}}{2}\cdot J_{-\frac{1}{2}}(at)\cdot u(t)<br />

    <br />
L^{-1}[F(s)]\,=-\,\frac{a^{\frac{1}{2}}}{2}\cdot J_{-\frac{1}{2}}(at)\cdot \frac{u(t)}{t}<br />
    wow, I don't know the inverse Laplace Transform can be written in terms of Bessel Fucntion I have another solution but your answer should be correct ,

    since
     J_{-\frac{1}{2}} (at) = \sqrt{\frac{2}{\pi a t}} \cos(at)
    the answer is jsut the function of cosine and t
    but what is  u(t) ?
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  14. #44
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    Quote Originally Posted by simplependulum View Post
    wow, I don't know the inverse Laplace Transform can be written in terms of Bessel Fucntion I have another solution but your answer should be correct ,

    since
     J_{-\frac{1}{2}} (at) = \sqrt{\frac{2}{\pi a t}} \cos(at)
    the answer is jsut the function of cosine and t
    but what is  u(t) ?

     u(t) is the Heaviside or unit step function, see here:
    Heaviside Step Function -- from Wolfram MathWorld
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  15. #45
    MHF Contributor Amer's Avatar
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    What is the new integral ?? luobo you should post an integral .
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