# Improper Integrals

• Feb 12th 2011, 12:27 PM
electricalphysics
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 $\displaystyle \displaystyle = Bln(b/a)- \int_{Ta}^{Tb} lnx(g'(x))dx$

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&#174; - Online Photo and Video Hosting

ImageShack&#174; - Online Photo and Video Hosting
• Feb 12th 2011, 02:39 PM
Random Variable
I don't know if this will help, but from the definiton of g(x), $\displaystyle \displaystyle g'(x) = - \frac{1}{x^{2}} \int_{1}^{x} f(t) \ dt + \frac{1}{x} \ f(x)$
• Feb 12th 2011, 05:28 PM
electricalphysics
Quote:

Originally Posted by Random Variable
I don't know if this will help, but from the definiton of g(x), $\displaystyle \displaystyle g'(x) = - \frac{1}{x^{2}} \int_{1}^{x} f(t) \ dt + \frac{1}{x} \ f(x)$

using this substitution i get
$\displaystyle \displaystyle = Bln(b/a)- \int_{Ta}^{Tb} lnx\frac{1}{x} (g(x)-f(x))dx$
which doesnt really get me anywhere, unless I'm missing something.
• Feb 12th 2011, 05:33 PM
zzzoak
$\displaystyle \displaystyle { I \: = \: \lim_{a,b \to \infty} \; \int_a^b \; lnx \: d(g(x))= \lim_{a,b \to \infty} \; ln \xi \: (g(b)-g(a)) \: \leq \: lnb \: |g(b)-g(a)| }$

As

$\displaystyle \displaystyle { \lim_{x \to \infty} \: g(x)=B }$

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

and we get

$\displaystyle | \: I \: |<\epsilon.$
• Feb 12th 2011, 11:07 PM
electricalphysics
Quote:

Originally Posted by zzzoak
$\displaystyle \displaystyle { I \: = \: \lim_{a,b \to \infty} \; \int_a^b \; lnx \: d(g(x))= \lim_{a,b \to \infty} \; ln \xi \: (g(b)-g(a)) \: \leq \: lnb \: |g(b)-g(a)| }$

As

$\displaystyle \displaystyle { \lim_{x \to \infty} \: g(x)=B }$

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

and we get

$\displaystyle | \: I \: |<\epsilon.$

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?
• Feb 13th 2011, 01:00 AM
zzzoak
$\displaystyle \displaystyle Bln(b/a)- \int_{Ta}^{Tb} lnx \: d(g(x))$

At limit the second term goes to zero and the first term is left.
• Feb 13th 2011, 11:47 AM
electricalphysics
Quote:

Originally Posted by zzzoak
$\displaystyle \displaystyle Bln(b/a)- \int_{Ta}^{Tb} lnx \: d(g(x))$

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 $\displaystyle \displaystyle ln \xi \: (g(b)-g(a))$, is this a sort of change of variables?
• Feb 13th 2011, 12:43 PM
zzzoak
It is the usage of the mean value theorem for integrals

$\displaystyle \int_a^b f(x)dx=f(c)\ \int_a^b dx=f(c)(b-a)$
• Feb 13th 2011, 02:18 PM
electricalphysics
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

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

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

$\displaystyle \displaystyle { = Bln(a/b)+\lim_{t \to \infty} (ln(c)(g(b)-g(a))$
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!
• Feb 15th 2011, 12:09 AM
zzzoak
$\displaystyle \displaystyle \int_1^{\infty} \frac {f(ax)}{x}dx=\int_a^{\infty} \frac {f(t)}{t}dt$

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

As
$\displaystyle 0<a<b$

$\displaystyle \displaystyle \int_a^{\infty}-\int_b^{\infty}= \int_a^{b}+ \int_b^{\infty}-\int_b^{\infty}=\int_a^{b} \; \; \mathbf{(1)}$

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:
$\displaystyle T_a<T_b, \ and \ T_a, \ T_b \to \infty$

$\displaystyle \displaystyle { \int_a^{\infty}-\int_b^{\infty}= \int_a^{b}+\int_b^{T_a}+ \int_{T_a}^{\infty}- \left( \; \int_b^{T_a}+\int_{T_a}^{T_b}+\int_{T_b}^{\infty} \left) =\int_a^{b}-\int_{T_a}^{T_b}+\left( \; \int_{T_a}^{\infty}-\int_{T_b}^{\infty} \ \left) }$

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

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