Deciding if Divergent/Convergent via Comparison Theorem

[drofenaz]

Homework Statement

Use the Comparison Theorem to determine whether the integral is convergent or divergent.
$\int_{0}^{1}\frac{e^{-x}}{\sqrt{x}}$

Homework Equations

The Attempt at a Solution

$\int_{0}^{1}\frac{e^{-x}}{\sqrt{x}} \leq \int_{0}^{1}\frac{1}{x^{\frac{1}{2}}}$

Because p<1, the integral is divergent. Wolfram, however, is giving me a finite answer, meaning the integral is actually convergent. Where am I going wrong?

 

[Mathoholic!]

The improper integral of (1/x^.5) is divergent. When you calculate its integral within a closed interval [0,a] you get a finite answer.

You're actually doing a reasonable comparison because (e^-x/x^.5) will (for x≥0, real numbers) always be under (1/x^.5), therefore, the integral between 0 and a (in this case, a=1) will always be less then the other, since it's the defined area between the graphs and the x axis.

Hence, you can conclude it's convergent.

 

[drofenaz]

The improper integral of (1/x^.5) is divergent. When you calculate its integral within a closed interval [0,a] you get a finite answer.

You're actually doing a reasonable comparison because (e^-x/x^.5) will (for x≥0, real numbers) always be under (1/x^.5), therefore, the integral between 0 and a (in this case, a=1) will always be less then the other, since it's the defined area between the graphs and the x axis.

Hence, you can conclude it's convergent.

I'm not sure I quite understand. When the comparison equation is near 0, the graph is up at +∞. So how is it possible that it is convergent?

 

[Mathoholic!]

The improper integral (as the endpoint of the interval [0,b] approaches ∞) is divergent. Meaning that if you calculate the area below the graph in this interval [0,∞[ you don't get a finite answer.
However the improper integral (regarding the startpoint of the interval [0,b]) is convergent, as weird as it may seem, because we're talking about the sqrt(x)^-1. If instead of having that you had x^-1 (they both have alike graphs), the same improper integral would give ∞ at x→0.

It's really interesting to see that expressions with basically the same graph have different responses in what concerns limits/integrals.

 

[drofenaz]

The improper integral (as the endpoint of the interval [0,b] approaches ∞) is divergent. Meaning that if you calculate the area below the graph in this interval [0,∞[ you don't get a finite answer.
However the improper integral (regarding the startpoint of the interval [0,b]) is convergent, as weird as it may seem, because we're talking about the sqrt(x)^-1. If instead of having that you had x^-1 (they both have alike graphs), the same improper integral would give ∞ at x→0.

It's really interesting to see that expressions with basically the same graph have different responses in what concerns limits/integrals.

So it's because of the limits that it is convergent. If it were from 1 to 0 it would be divergent?

 

[Mark44]

So it's because of the limits that it is convergent. If it were from 1 to 0 it would be divergent?

No. Changing the limits of integration just changes the sign of the value of the integral.

Assuming that f is integrable on [a, b],

$$\int_a^b f(x)dx = -\int_b^a f(x)dx$$

 

[drofenaz]

I was mixing up my p-test for 0->∞ and 0->1.