Can a substitution cause a convergent definite integral to diverge?

Is it possible for an integral, $$\int_a^b f(x)\text{d}x,$$ to converge, but for some substitution $x=g(y)$ to cause the integral, $$ \int_{g^{-1}(a)}^{g^{-1}(b)} f(g(y)) g'(y) \text{d}{y},$$ to diverge?

I suppose this could happen if $$f(g(y))g'(y)\longrightarrow\infty$$ as $y\longrightarrow g^{-1}(a)$ or $y\longrightarrow g^{-1}(b)$ ? Or, equivalently, if $$f(a)g'(g^{-1}(a))=\infty$$ or $$f(b)g'(g^{-1}(b))=\infty.$$

Edit

I originally asked this question because certain substitutions appeared to make the integrand become "pathalogical" or to blow up to infinity at certain points, so from an "area under the curve" point of view I found it difficult to see how convergence could persist. However, indeed such functions can converge, e.g. $ x^{-1/2} $ over $[0,1] $

Solution 1:

Consider the case of an non closed interval.

$$\int_a^b f(x) dx = \lim_{a'\downarrow a, b'\uparrow b} \int_{a'}^{b'}f(x)dx$$ ans the integrals $$ \int_{a'}^{b'}f(x)dx $$are definite Riemann integrals (so $f$ is bounded on $[a',b']$).

Now make the change in the definite integrals: $$ \int_{a'}^{b'}f(x)dx = \int_{g^{-1}(a')}^{g^{-1}(b')} f(g(y)) g'(y) {d}{y} $$ These integrals are well defined. Now the first integral si convergent iff $$ \lim_{a'\downarrow a, b'\uparrow b} \int_{a'}^{b'}f(x)dx = \lim_{a'\downarrow a, b'\uparrow b} \int_{g^{-1}(a')}^{g^{-1}(b')} f(g(y)) g'(y) {d}{y} $$ exists. As $g^{-1}$ is continuous when $g\in C^1$, the preceding expression becomes (assuming $g$ increasing, otherwise just change the inequalities accordingly):

$$ \lim_{ga\downarrow g^{-1}(a), gb\uparrow g^{-1}(b)} \int_{ga}^{gb} f(gy)) g'(y) {d}{y} $$

which proves that one integral is convergent iff the other one is.