Does the following integral converge

[a1010711]

Homework Statement

the integral from 0 to 1 given:

1 / [ (x^1/3)(x^2+2x)^1/2 ] dx

please explain,, thank you!

The Attempt at a Solution

At 0 the integrand behaves as 1/x^1/3*x^1/2 = 1/x^1/6 which is convergent as the exponent is <1

 

[Mark44]

Homework Statement

the integral from 0 to 1 given:

1 / [ (x^1/3)(x^2+2x)^1/2 ] dx

please explain,, thank you!

The Attempt at a Solution

At 0 the integrand behaves as 1/x^1/3*x^1/2 = 1/x^1/6 which is convergent as the exponent is <1

The integrand does not behave as 1/[x^(1/3)*x^(1/2)]. Even if it did, 1/x^(1/6) becomes infinitely large as x approaches zero. You're probably thinking of the behavior of a p series, where n is growing infinitely large.

My guess is that this is a divergent integral because of what's happening close to zero. The integrand is less than 1/x^(4/3), which is a function I can integrate and for which the definite integral diverges. Unfortunately, our integrand is less than a function whose antiderivative diverges, so that's no help.

On the other hand, x^(1/3)*sqrt(x^2 + 2x + 1) > x^(1/3)*sqrt(x^2 + 2x), so
1/[x^(1/3)*sqrt(x^2 + 2x + 1)] < 1/[x^(1/3)*sqrt(x^2 + 2x)] ,
which means that
1/[x^(1/3)*(x + 1)] < 1/[x^(1/3)*sqrt(x^2 + 2x)]

If I can show that the antiderivative of the function on the left above is divergent, that means that the one on the right is, too. Unfortunately, now, I can't come up with an antiderivative for the function on the left.

 

[lanedance]

could maybe do it 2 ways based on the integral 1->inf of 1/x^n converging iff n>1

2 ways potentially

substitute straight into your integral, changing integration from 0to1 to 1 to infinty, and then use some comparison theorems

or substitute into the integral 1->inf of 1/x^n which will change it to 0 to 1, should give you a condition based on n for this integral to converge

i'm thinking it probably does converge at this stage

 

[Mark44]

could maybe do it 2 ways based on the integral 1->inf of 1/x^n converging iff n>1

2 ways potentially

substitute straight into your integral, changing integration from 0to1 to 1 to infinty, and then use some comparison theorems

Substitute WHAT into the integral?

or substitute into the integral 1->inf of 1/x^n which will change it to 0 to 1, should give you a condition based on n for this integral to converge

i'm thinking it probably does converge at this stage

Lanedance, I'm not following you here.

 

[lanedance]

woops sorry missed out the key part... editing before submitting then not reading properly

meant to say substitute u = 1/x, into either integral to change to zero limit to a limit at infinity, or vice versa

so du = -1/x^2 dx ---> -du/u^2 = x