How to Solve Improper Integral \int_{0}^{2} \frac{1}{1-x^{1/3}} dx

[PCSL]

[tex]\int_{0}^{2} \frac{1}{1-x^{1/3}} dx[/tex]
I then would break up into:
[tex]\int_{0}^{1} \frac{1}{1-x^{1/3}} dx + \int_{1}^{2} \frac{1}{1-x^{1/3}} dx[/tex]

I'm lost on where to go from here. The integral looks so simple but I'm not sure if I should make a u sub or a trig sub. Could someone give me a hint on how to solve it?

 

[Harrisonized]

You don't need to break it up. Use the u-sub u=x^1/3.

 

[PCSL]

You don't need to break it up. Use the u-sub u=x^1/3.

Why wouldn't you split it up if x is unbounded at 1?

 

[dynamicsolo]

Here's a small hint for dealing with such situations: keep in mind that the "1" in the numerator is also "x⁰"...

 

[Harrisonized]

Why wouldn't you split it up if x is unbounded at 1?

It cancels out unless the limit of the integral doesn't exist there.

 

[PCSL]

Thanks dynamic, but I still have no idea where to start. and what cancels out Harris?

 

[dynamicsolo]

If u = 1 - x^1/3 , what is du? You've taken care of the original denominator then, but picked up something new down there. What would go in the numerator to get rid of it? You should have an integral in u that you can deal with. (This is called a "rationalizing substitution" by some authors.)

Don't forget to transform your limits in the definite integral if you finish the integral in u without transforming back to x.

 

[PCSL]

If u = 1 - x^1/3 , what is du? You've taken care of the original denominator then, but picked up something new down there. What would go in the numerator to get rid of it? You should have an integral in u that you can deal with. (This is called a "rationalizing substitution" by some authors.)

Don't forget to transform your limits in the definite integral if you finish the integral in u without transforming back to x.

Alright, one more question. Is there any shortcut to finding out of a function diverges or converges or do you have to wait until you get to the end of solving the equation.

 

[Harrisonized]

what cancels out Harris?

∫_a^c f'(x) dx
= ∫_a^b f'(x) dx + ∫_b^c f'(x) dx
= [f(b)-f(a)] + [f(c)-f(b)]
= f(c)-f(a)

Usually, you split integrals up when lim f(x) as x→b doesn't exist, but lim f(x) as x→b⁺ and lim f(x) as x→b^- do exist and they're not the same. Then the integral becomes:

∫_a^c f'(x) dx
= ∫_a^b- f'(x) dx + ∫_b+^c f'(x) dx
= [f(b^-)-f(a)] + [f(c)-f(b⁺)] (Keep in mind that f(b^-) and f(b⁺) are limits.)

Alright, one more question. Is there any shortcut to finding out of a function diverges or converges or do you have to wait until you get to the end of solving the equation.

In this case, you should wait for the end. Since the indefinite integral can be found, it's much easier to wait for the end.

 

[dynamicsolo]

In the case of an integral like this, there is something you can look at. As x approaches 1 "from the left", the integrand runs off to "positive infinity"; however, as x approaches 1 "from the right", the integrand goes to "negative infinity". Most authors I've run into reject the total as a convergent sum, even if the individual integrals converge (but they don't for this integral); many physicists might say otherwise...

There are things like the "p-test" for convergence/divergence, which can be adapted to Type II improper integrals like the one you're working on. For this particular integrand, we have to be somewhat careful, since it doesn't have quite the right form; the treatment is probably a bit beyond what you'll see in your course. In any case, for now, you'll be expected to show the full-out integration in homework or exams.

(Let's say I wasn't surprised to find that this integral diverges...)

 

[Harrisonized]

(Let's say I wasn't surprised to find that this integral diverges...)

I'm going to disagree here. This might not sound very mathematical or rigorous, but let's consider the function f'(x)=1/x. From a geometric standpoint,

∫_-1¹ 1/x dx = 0

because the function is symmetrical and it just so happens that

∫_-b^-a 1/x dx
= -∫_a^b 1/x dx

The function in the original post might not be symmetrical, but I think the same logic applies.

 

[dynamicsolo]

I'm going to disagree here. This might not sound very mathematical or rigorous, but let's consider the function f'(x)=1/x. From a geometric standpoint,

∫_-1¹ 1/x dx = 0

because the function is symmetrical and it just so happens that

∫_-b^-a 1/x dx
= -∫_a^b 1/x dx

The function in the original post might not be symmetrical, but I think the same logic applies.

I'll say, go argue with the analysts. Many physicists will say the "opposing infinities" should cancel, but books I've dealt with would reject, for instance, even [itex] \int_{-a}^{a} \frac{dx}{x^{1/3}} [/itex] as being convergent...

 

[Dick]

I'll say, go argue with the analysts. Many physicists will say the "opposing infinities" should cancel, but books I've dealt with would reject, for instance, even [itex] \int_{-a}^{a} \frac{dx}{x^{1/3}} [/itex] as being convergent...

There's really no argument. Evaluating the integral outside of a symmetrical interval around the singularity and letting the interval go to zero gives you the Cauchy principal value of the integral. It's useful for some problems like in distribution theory. But the integral in the mathematical sense of Riemann or Lebesgue (where you have to be able to sum in any possible way, not just symmetrically) still doesn't exist.

 

[PCSL]

If I find [tex]ln(0)[/tex] when evaluating the limit would that mean the integral is diverging.

 

[dynamicsolo]

Yes, it does.