Convergent or Divergent: Integral of Square Root of t Times Exponential Function

[Justabeginner]

Homework Statement

Does the following integral converge or diverge?
$\int_{0}^{∞} √t * e^{-t}\, dt$
Exponential function is not under the square root symbol.

Homework Equations

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!

 

[voko]

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

 

[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.

 

[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.

 

[SteamKing]

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

 

[LCKurtz]

Homework Statement

Does the following integral converge or diverge?
$\int_{0}^{∞} √t * e^{-t}\, dt$
Exponential function is not under the square root symbol.

Homework Equations

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!

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

 

[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?

 

[voko]

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

 

[Justabeginner]

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

sqrt(t) < t ?

 

[LCKurtz]

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

sqrt(t) < t ?

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

 

[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.

 

[voko]

Yes. So an integral test is essentially a comparison test but for integrals.

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.

 

[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.

 

[Justabeginner]

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

 

[LCKurtz]

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

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.

 

[Justabeginner]

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

 

[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.

 

[voko]

Yes, that will work.

 

[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.