Does the following integral converge

a1010711

1. The problem statement, all variables and given/known data

the integral from 0 to 1 given:

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

plz explain,, thank you!

3. 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

Mentor
1. The problem statement, all variables and given/known data

the integral from 0 to 1 given:

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

plz explain,, thank you!

3. 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

Homework Helper
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

Mentor
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

Homework Helper
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

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