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Math Help - Improper Integrals

  1. #1
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    Improper Integrals

    Hello

    I'm having difficulty with part (ii) of the first link.

    I used the result from part (i) and was able to get g(Tb)-g(Ta)- integral from Ta to Tb of g(x)/x dx...

    I noticed that the g(Ta)-g(Tb) terms cancel and i'm only left with the integral term.
    I used by parts on the integral term with u=g(x) , du= g'(x) dv= 1/xdx, v= lnx.
    I obtained  \displaystyle = Bln(b/a)-  \int_{Ta}^{Tb}  lnx(g'(x))dx


    My problem is that I can't seem to get rid of the final integral term. Any help would greatly be appreciated.

    I'm also stuck with the question in the second image, I believe that this questions requires me to apply my result from part(ii)?

    Thank you.

    Question part(ii) ImageShack® - Online Photo and Video Hosting

    ImageShack® - Online Photo and Video Hosting
    Last edited by electricalphysics; February 12th 2011 at 08:03 PM.
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  2. #2
    Super Member Random Variable's Avatar
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    I don't know if this will help, but from the definiton of g(x),  \displaystyle g'(x) = - \frac{1}{x^{2}} \int_{1}^{x} f(t) \ dt + \frac{1}{x} \ f(x)
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  3. #3
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    Quote Originally Posted by Random Variable View Post
    I don't know if this will help, but from the definiton of g(x),  \displaystyle g'(x) = - \frac{1}{x^{2}} \int_{1}^{x} f(t) \ dt + \frac{1}{x} \ f(x)
    using this substitution i get
     \displaystyle = Bln(b/a)-  \int_{Ta}^{Tb}  lnx\frac{1}{x} (g(x)-f(x))dx
    which doesnt really get me anywhere, unless I'm missing something.
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  4. #4
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    <br />
\displaystyle {<br />
I \: = \: \lim_{a,b \to \infty} \; \int_a^b \; lnx \: d(g(x))= \lim_{a,b \to \infty} \; ln \xi \: (g(b)-g(a)) \: \leq \: lnb  \: |g(b)-g(a)|<br />
}<br />

    As

    <br />
 \displaystyle {<br />
\lim_{x \to \infty} \: g(x)=B<br />
}<br />

    <br />
\displaystyle {<br />
|g(b)-g(a)| \: \leq \: |g(b)-B|+|g(a)-B|  \:< \: \frac{\epsilon}{lnb}<br />
}<br />

    and we get

    <br />
| \: I \: |<\epsilon.<br />
    Last edited by zzzoak; February 13th 2011 at 01:56 AM.
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  5. #5
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    Quote Originally Posted by zzzoak View Post
    <br />
\displaystyle {<br />
I \: = \: \lim_{a,b \to \infty} \; \int_a^b \; lnx \: d(g(x))= \lim_{a,b \to \infty} \; ln \xi \: (g(b)-g(a)) \: \leq \: lnb  \: |g(b)-g(a)|<br />
}<br />

    As

    <br />
 \displaystyle {<br />
\lim_{x \to \infty} \: g(x)=B<br />
}<br />

    <br />
\displaystyle {<br />
|g(b)-g(a)| \: < \: \frac{\epsilon}{lnb}<br />
}<br />

    and we get

    <br />
| \: I \: |<\epsilon.<br />
    Thank you very much for helping me,

    but I dont see what showing that this limit is less then epsilon allows me to do to evaluate the integral?
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  6. #6
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    <br />
\displaystyle  Bln(b/a)- \int_{Ta}^{Tb} lnx \: d(g(x))<br />

    At limit the second term goes to zero and the first term is left.
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  7. #7
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    Quote Originally Posted by zzzoak View Post
    <br />
\displaystyle  Bln(b/a)- \int_{Ta}^{Tb} lnx \: d(g(x))<br />

    At limit the second term goes to zero and the first term is left.
    Thank you very much, it's very appreciated. I do understand your logic but I dont understand how the integral becomes \displaystyle   ln \xi \: (g(b)-g(a)) , is this a sort of change of variables?
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  8. #8
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    It is the usage of the mean value theorem for integrals

    <br />
\int_a^b  f(x)dx=f(c)\ \int_a^b  dx=f(c)(b-a)<br />
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  9. #9
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    Thank you very much, I was able to complete part(ii).

    I'm now a little stuck on part(iii),

    I was able to seperate the integral and use change of variables where u=ax and z=bx.
    From there

    <br />
\displaystyle {<br />
= Bln(a/b)+\lim_{t \to \infty} (\; \int_a^t \; ln(u)((g'(u))du +\; \int_b^t \; ln(z)((g'(z))dz )}

    Then applying the mean value theorem for integrals from similarily from part (ii),

    <br />
\displaystyle {<br />
= Bln(a/b)+\lim_{t \to \infty} (ln(c)(g(b)-g(a))
    where c belongs to [a,b]

    I'm stuck at this step, I'm having a hard time relating the 2nd term to the required f(t)/t integral.

    Thanks!
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  10. #10
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    <br />
\displaystyle<br />
\int_1^{\infty} \frac {f(ax)}{x}dx=\int_a^{\infty} \frac {f(t)}{t}dt<br />

    <br />
\displaystyle<br />
\int_1^{\infty} \frac {f(ax)-f(bx)}{x}dx=\int_a^{\infty} \frac {f(t)}{t}dt-\int_b^{\infty} \frac {f(t)}{t}dt<br />

    As
    <br />
0<a<b<br />

    <br />
\displaystyle<br />
\int_a^{\infty}-\int_b^{\infty}= \int_a^{b}+ \int_b^{\infty}-\int_b^{\infty}=\int_a^{b} \; \; \mathbf{(1)} <br />

    It seems that the division of integral into two is not true for indefinite integrals. We get
    only one term in (iii).

    So we rewrite it:
    <br />
T_a<T_b, \ and \ T_a, \ T_b \to \infty<br />

    <br />
\displaystyle {<br />
\int_a^{\infty}-\int_b^{\infty}= \int_a^{b}+\int_b^{T_a}+ \int_{T_a}^{\infty}- \left( \;  \int_b^{T_a}+\int_{T_a}^{T_b}+\int_{T_b}^{\infty}  \left) =\int_a^{b}-\int_{T_a}^{T_b}+\left( \;   \int_{T_a}^{\infty}-\int_{T_b}^{\infty} \ \left)  <br />
}<br />

    From (ii)
    <br />
\displaystyle {<br />
lim_{T_a, \ T_b \to \infty} \int_{T_a}^{T_b}=B \  ln(T_b/T_a)<br />
}<br />

    May be smb could explain why (1) is not true. Thanks.
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