Convergence of Improper Integral

In summary, the original integral does not converge due to the behavior of the integrand as x approaches 1.
  • #1

pbialos

I have to analize the convergence of the following integral:

[tex]\int_0^1 \frac {x^2+1} {\sqrt x * (1-x)^{5/4}}[/tex]

I tried to divide it between 0-1/2 and 1/2-1 and on the first one i reached to:
[tex]\int_0^{1/2} \frac {x^2+1} {\sqrt x * (1-x)^{5/4}}<=\int_0^{1/2} \frac {x^2+1} {x^{14/4}}[/tex]
can i say that this integral converges and therefore the orgininal converges?, and more important, how would i justify that the last integral converges in an exam?
please correct any mistakes that i probably had made, and forgive me for me awful english.

Many Thanks, Paul.
 
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  • #2
How did you get that "14/4". If you just add the powers in the denominator you get 13/4. Are you arguing that [tex]\frac{x^2+1}{x^{\frac{13}{4}}}< \frac{x^2+1}{x^{\frac{14}{4}}}?

In any case, you can't just add the exponents like that. If x is close to 0, then
1- x is close to 1 and x2+1 is close to 1. For x close to 0, the integrand is close to [tex]\frac{1}{\sqrt{x}}= x^{-\frac{1}{2}}[/tex]. And, of course, that integral exists.

To look at what happens for x close to 1, it might help to make the substitution
u= 1- x. Then x2+ 1= 2- 2u+ u2 and the integrand becomes [tex]\frac{2- 2u+ u}{\sqrt{1-u}u^{\frac{5}{4}}[/tex]. What is that like when u is close to 0?
 
  • #3
The first integrand varies as 1/sqrt(x) near zero ( I checked what the expression becomes for small values of x by means of the power series expansion.) .So, the integral will converge or diverge according as the integral of 1/sqrt(x).It converges.
Similarly,to put in awful technical language, the integrand becomes
(1^2 +1) /sqrt(1)*( 1-x)^5/4 near 1 & diverges because the integral
dx/(1-x)^5/4 diverges if taken over a neighbourhood of 1.
These ideas must be put in terms of inequalities; e.g. in the first case, constants p & q could be found out such that p/sqrt(x) < the integrand < q/sqrt(x)
holds in a neighbourhood of zero.
I am, with great respect,
Einstone.
 
  • #4
Don't know what happened to my first response- and no "edit" button to fix it!

What I was saying is that since x2+1 and 1- x are both close to 1 for x close to 0, the integrand is close to x-1/2 which is integrable.

If you let u= 1- x, then x= 1- u so x2+1= u2- 2u+ 1 and the integrand becomes [tex]\frac{u^2- 2u+ 1}{\sqrt{1-u}u^\frac{5}{4}}[/tex].
For u close to 1 that is close to [tex]\frac{1}{u^\frac{5}{4}}[/tex] which does not converge.
 
  • #5
thank you

First of all, thank you for your prompt and clear reply. I think i now understand how to do this exercise, but i don't know how to explain on a test why i can replace the integrands with another integrand that behaves the same way when x is close to 0 in the first part of the integral for example. I really don't understand the part of your explanation where you say that it can be found two constants p and q such that p/sqrt(x)<integrand<q/sqrt(x). I think that you are trying to show that i can replace the integrand by 1/sqrt(x)
but i don't know if it would be enough justification on a test.
Many Thanks, Paul.
 
  • #6
Would this be sufficient:

[tex]\frac{x^2+1}{\sqrt{x}(1-x)^{5/4}}>\frac{1}{\sqrt{x}(1-x)^{5/4}}>\frac{1}{\sqrt{x}(1-x)^2}[/tex]

for all [itex]x\in (0,1)[/tex]

Making the substitution [itex]u=\sqrt{x}[/itex] and using partial fractions, we obtain:

[tex]\int \frac{dx}{\sqrt{x}(1-x)^2}=\frac{1}{2}\left[\frac{1}{1-\sqrt{x}}-ln(1-\sqrt{x})-\frac{1}{1+\sqrt{x}}+ln(1+\sqrt{x})\right][/tex]

which diverges as x goes to 1.

Thus:

[tex]\int_0^1 \frac{x^2+1}{\sqrt{x}(1-x)^{5/4}}[/tex]

also diverges.
 

1. What is an improper integral?

An improper integral is a type of integral in calculus where either the upper or lower limit of integration is infinite or the integrand function is unbounded at one or more points within the specified interval.

2. How is an improper integral calculated?

An improper integral is calculated by taking the limit as one or both of the limits of integration approach infinity or a point where the function is unbounded. This limit is then evaluated to determine the convergence or divergence of the integral.

3. What is the difference between a convergent and divergent improper integral?

A convergent improper integral is one where the limit of the integral exists and is a finite number. This means that the integral is well-defined and can be calculated. A divergent improper integral is one where the limit does not exist, meaning that the integral is undefined and cannot be calculated.

4. How can I determine the convergence or divergence of an improper integral?

The convergence or divergence of an improper integral can be determined by evaluating the limit as the limits of integration approach infinity or a point of unboundedness. If the limit exists and is a finite number, the integral is convergent. If the limit does not exist, the integral is divergent.

5. What are some common applications of improper integrals in science?

Improper integrals are used in many areas of science, including physics, engineering, and economics. They are particularly useful in calculating quantities that are infinite or unbounded, such as the area under a curve that extends to infinity or the volume of a solid with an unbounded region. Improper integrals are also used in solving differential equations and in probability and statistics.

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