Does This Improper Integral Converge or Diverge?

[crowKAKAWWW]

Hi, I need to determine whether this improper integral converges or diverges

<br /> <br /> \int_{-1}^{1} \frac{x}{\sqrt{1-x^2}}dx<br /> <br />


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

\int_{-1}^{0} \frac{x}{\sqrt{1-x^2}}dx \ + \ \int_{0}^{1} \frac{x}{\sqrt{1-x^2}}dx


I've integrated it and got

<br /> <br /> - \frac{1}{2\sqrt{1-x^2}}<br /> <br />


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?

 

[Mark44]

Hi, I need to determine whether this improper integral converges or diverges

<br /> <br /> \int_{-1}^{1} \frac{x}{\sqrt{1-x^2}}dx<br /> <br />


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

\int_{-1}^{0} \frac{x}{\sqrt{1-x^2}}dx \ + \ \int_{0}^{1} \frac{x}{\sqrt{1-x^2}}dx


I've integrated it and got

<br /> <br /> - \frac{1}{2\sqrt{1-x^2}}<br /> <br />


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.

 

[crowKAKAWWW]

Ah ok,

So because of the denominator

<br /> \sqrt{1-x^2}}<br />

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?

 

[JThompson]

I've integrated it and got

<br /> <br /> - \frac{1}{2\sqrt{1-x^2}}<br /> <br />

Just to be clear, the antiderivative of

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

is not
- \frac{1}{2\sqrt{1-x^2}}.


It is -\sqrt{1-x^{2}}.

 

[Mark44]

Ah ok,

So because of the denominator

<br /> \sqrt{1-x^2}}<br />

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.