# Integral-Converge or Diverge?

1. Aug 4, 2013

### Justabeginner

1. The problem statement, all variables and given/known data
Does the following integral converge or diverge?
$\int_{0}^{∞} √t * e^{-t}\, dt$
Exponential function is not under the square root symbol.

2. Relevant equations

3. The attempt at a solution
Using the ratio test, I have:

sqrt(t+1)* e^(-{t+1})/[sqrt(t)* e^(-t)] = lim t-> ∞ sqrt(t+1)/(e*sqrt(t)) = 0
0 < 1
Therefore, I think the series is absolutely convergent.
Is this correct, and is the logic and method I used correct? Thank you!

Last edited: Aug 4, 2013
2. Aug 4, 2013

### voko

The ratio test is for series, not for integrals. How can you use it here?

3. Aug 4, 2013

### Justabeginner

That was a stupid mistake. I shall redo the problem using the integral test:

v= sqrt(t)
t= v^2
dt= 2v dv
1/2 ∫v*e^(-v^2) dv since u=-v^2 and 0 and ∞ are limits
1/2 ∫v*e^(-v^2) dv with 0 and z being limits of integration
lim z-> ∞ \left.\frac{-1}{2}*e^(-v^2)\right|_0^z = lim z-> ∞ (1/2 - 1/2*e^(-z^2))= 1/2

Therefore, if this integral converges, series also converges by the Integral Test.

Is this logic accurate? Thanks.

Last edited: Aug 4, 2013
4. Aug 4, 2013

### voko

If I understand your intent correctly, you want to reduce the analysis of the integral's convergence to a series' convergence, and you do that via the integral test.

That is an acceptable strategy. But I do not understand at all what you do. I see that you have a substitution, but I do not see how that is related to the integral test. The integral test has a requirement for the integrand, and I would expect that you start by checking that requirement.

5. Aug 4, 2013

### SteamKing

Staff Emeritus
e^(-t) is not the natural log function. e^x is the exponential function.

6. Aug 4, 2013

### LCKurtz

Hint: On $[1,\infty)$ how do $\sqrt t$ and $t$ compare?

7. Aug 4, 2013

### Justabeginner

Thank you SteamKing. I've edited that.

And I thought the integral test involves picking out a known function, determining convergence for that series, and then comparing the two series?

8. Aug 4, 2013

### voko

The integral test establishes a link between the convergence of an integral and the convergence of a series. There are no "two" series.

9. Aug 4, 2013

### Justabeginner

sqrt(t) < t ?

10. Aug 4, 2013

### LCKurtz

Right. Do you see how to make a comparison using that fact and making an easier integral to work?

11. Aug 4, 2013

### Justabeginner

Yes. So an integral test is essentially a comparison test but for integrals. In this case, I would be able to use the comparison between sqrt(t) and t? I didn't know that-I presumed that I had to use something involving e.

12. Aug 4, 2013

### voko

The integral test is a test for series, involving integrals. It can also be used as a test for integrals involving series. It is not a comparison test.

13. Aug 4, 2013

### LCKurtz

Think about what happens if you replace the $\sqrt t$ in the integral with $t$. You can't ignore the exponential. And you need to say something about the different interval.

14. Aug 4, 2013

### Justabeginner

I thought the sqrt(t) just minimizes the function (decreases values)? So the exponential wouldn't matter either way?

15. Aug 4, 2013

### LCKurtz

Assuming nonnegative integrand, if you want to show convergence, you have to compare the integral you are given with a new integral that is greater and whose integral converges.

16. Aug 4, 2013

### Justabeginner

So could I pick the new integral as t * e^t ? That gives me greater values, I think?

17. Aug 4, 2013

### Justabeginner

That integral is wrong (does not converge- increases without bound, I think). Can I try this one?

(t) e^(-t)
And the limits of integration should be 1 to infinity, I think? That way, the first integral will be <= the second integral.

18. Aug 4, 2013

### voko

Yes, that will work.

19. Aug 4, 2013

### LCKurtz

You have a couple of details left. Your original was$$\int_0^\infty \sqrt t e^{-t}\, dt$$You can't compare it directly with $$\int_1^\infty te^{-t}\, dt$$because they aren't over the same interval. You have to address that, even if it is easy.