Convergence of an Integral

The method that you will use to evaluate this integral is similar to the one used to evaluate the integral \int_{1}^{+infty}\frac{1}{x^{2}}dxIn summary, the original function DNE at -1 and 1, so the limits need to be split into two integrals. After integrating, the value of the integral converges to 0, not -1 as previously thought. This can be determined by using limits to evaluate the integral, where one of the split integrals will converge to a value, and the other will converge to the negative of that value due to the integrand being an odd function. The antiderivative of \frac{x}{\sqrt{1-x^2
  • #1
crowKAKAWWW
2
0
Hi, I need to determine whether this improper integral converges or diverges

[tex]

\int_{-1}^{1} \frac{x}{\sqrt{1-x^2}}dx

[/tex]


The original function DNE at -1, 1 so I split the limits

[tex] \int_{-1}^{0} \frac{x}{\sqrt{1-x^2}}dx \ + \ \int_{0}^{1} \frac{x}{\sqrt{1-x^2}}dx [/tex]


I've integrated it and got

[tex]

- \frac{1}{2\sqrt{1-x^2}}

[/tex]


After that I tried subbing in the limits for each equation and added them together

-0.5 + -0.5 = -1

So my answer was that the integral converges to -1, but the answer is supposed to be 0, can anyone help me out with this?
 
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  • #2
crowKAKAWWW said:
Hi, I need to determine whether this improper integral converges or diverges

[tex]

\int_{-1}^{1} \frac{x}{\sqrt{1-x^2}}dx

[/tex]


The original function DNE at -1, 1 so I split the limits

[tex] \int_{-1}^{0} \frac{x}{\sqrt{1-x^2}}dx \ + \ \int_{0}^{1} \frac{x}{\sqrt{1-x^2}}dx [/tex]


I've integrated it and got

[tex]

- \frac{1}{2\sqrt{1-x^2}}

[/tex]


After that I tried subbing in the limits for each equation and added them together

-0.5 + -0.5 = -1

So my answer was that the integral converges to -1, but the answer is supposed to be 0, can anyone help me out with this?
You need to use limits to evaluate this integral. Splitting the integral into two integrals is a good idea. You can get away with evaluating just one of the improper integrals, because the integrand is an odd function. If one of your split integrals converges to a value, the other one will converge to the negative of that value.
 
  • #3
Ah ok,

So because of the denominator

[tex]
\sqrt{1-x^2}}
[/tex]

In order to get a real number... x^2 < 1
So the limits would be (-1, 1), added together gives me my convergence to 0, is that it?
 
  • #4
crowKAKAWWW said:
I've integrated it and got

[tex]

- \frac{1}{2\sqrt{1-x^2}}

[/tex]

Just to be clear, the antiderivative of

[tex] \frac{x}{\sqrt{1-x^2}}[/tex]

is not
[tex]- \frac{1}{2\sqrt{1-x^2}}[/tex].


It is [tex]-\sqrt{1-x^{2}}[/tex].
 
  • #5
crowKAKAWWW said:
Ah ok,

So because of the denominator

[tex]
\sqrt{1-x^2}}
[/tex]

In order to get a real number... x^2 < 1
So the limits would be (-1, 1), added together gives me my convergence to 0, is that it?
I think that you completely missed my point. Because the integrand is not defined at every point in the interval [-1, 1], you can't just find an antiderivative and evaluate it at the endpoints.

Your textbook should have some examples of how to work with improper integrals.
 

1. What is the definition of convergence of an integral?

The convergence of an integral is a mathematical concept that determines whether or not a certain integral (a mathematical expression used to calculate the area under a curve) has a finite value. In other words, it is a way to determine whether or not an integral "converges" or approaches a specific value as the limits of integration (the start and end points of the integral) approach infinity.

2. How is convergence of an integral related to series convergence?

The convergence of an integral is closely related to the convergence of a series (an infinite sum of terms). In fact, the convergence of an integral can be seen as a special case of series convergence. If an integral converges, then the corresponding series also converges, and vice versa.

3. What is the importance of convergence of an integral in calculus?

The concept of convergence of an integral is important in calculus because it allows us to determine whether or not certain integrals have a finite value. This is crucial in many applications of calculus, such as in physics and engineering, as it allows us to accurately calculate areas, volumes, and other quantities.

4. How do you determine the convergence of an improper integral?

An improper integral is an integral with one or both of its limits of integration being infinity or a point where the integrand (the function being integrated) is undefined. To determine the convergence of an improper integral, one must evaluate the integral as a limit, taking into account any discontinuities in the integrand. If the limit exists, then the integral is said to converge.

5. Can an integral converge even if its integrand is undefined at some point?

Yes, an integral can still converge even if its integrand is undefined at some point. This is because the value of the integral is determined by the behavior of the integrand as it approaches the point of discontinuity. As long as the limit of the integrand exists at the point of discontinuity, the integral can still converge.

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