College Math Teaching

February 19, 2012

Divergent Improper Integrals: change of variables to an unbounded integrand.

Filed under: calculus, change of variable, improper integrals, integrals, integration by substitution, pedagogy — collegemathteaching @ 10:41 pm

This post was motivated by a student question: my student wanted help with the following problem:

\int^{\infty}_{1} \frac{x^2}{\sqrt{x^3 +1}} dx
Of course the idea is to do a substitution: u = x^3 + 1 which transforms the integral into \frac{1}{3} \int^{\infty}_{2} \frac{1}{\sqrt{u}} du which diverges. So far, so good. But then I told him one of my calculus tips: “it is often a good idea to try to guess the answer ahead of time” and then pointed out that for large values of x, \frac{x^2}{\sqrt{x^3 +1}} \approx \frac{x^2}{\sqrt{x^3}} = \sqrt{x} and of course \int^{\infty}_1 \sqrt{x} dx diverges because the integrand does not go to zero (in fact, is unbounded!) as x tends to infinity.

Then I realized that a change of variables had taken an unbounded function to a bounded one…though one which did not produce a convergent improper integral.

That lead to the natural question: if one has an integrand which is positive but monotonically decreasing to zero on [1, \infty ) , is there a change of variables which will change the integrand to either an unbounded function on [1, \infty ) or at least one that does not decrease to zero?

I admit that I have not answered this question yet, nor have I looked it up. But I can answer this question for a certain class of functions:


Given \int^{\infty}_1 \frac{1}{x^r} dx
If 0 < r < 1 , let k > \frac{1}{1-r} . Then the change of variable u = x^{\frac{1}{k}} transforms \int^{\infty}_{1} \frac{1}{x^r} dx to k \int^{\infty}_1 u^{k(1-r) -1} du and of course u^{k(1-r) -1} is unbounded on [1, \infty) .

If 1 < r let k < \frac{1}{1-r} < 0 . Then \int^{\infty}_1 \frac{1}{x^r} dx is transformed into |k|\int^{1}_{0} u^{-1+k(1-r)} du which is an integral of a bounded function over a bounded region.

In short, one class of functions whose improper integral diverges can be transformed to functions that tend to infinity and the class of functions whose integrals converge can be transformed into functions which are bounded over a bounded interval.

Here is such an example: We show the equivalent integrals \int^{1.5}_{1} 3x^{\frac{1}{2}} dx and \int^{(1.5)^3}_1 \frac{1}{\sqrt{u}} du . The transformation is accomplished by using u =x^3 . Note how the transformation stretches the interval of integration to account for the function “shrinkage”.

On the other hand, using u^{-2}=x transforms \int^{\infty}_1 \frac{1}{x^2} dx into 2\int^{1}_{0} u du


  1. Integral of 1/lnx is monotonically decreasing, yet diverges when evaluated from 1 to infinity. Performing the substitution u=lnx transforms the integrand into (e^u)/u, which is clearly unbound asa u goes to infinity. Fairly certain this method is valid for any integrand of the form 1/(lnx)^p for p>1.

    Comment by John — February 8, 2018 @ 8:54 pm

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