# How can a doubly improper integral become a singly improper integral after substitution?

How can a doubly improper integral become a singly improper integral after substitution?

If we take $$x=\sin^2{t}$$ then:

$$\int_0^1\frac{1}{x\sqrt{1-x}}dx=2\int_0^{\pi/2}\csc{t}dt$$

On the lefthand side the integral is doubly improper on both the lower ($$0$$) and upper ($$1$$) bound. But after substitution the new integral is only improper at the lower bound ($$0$$). How come? What happened to the improper upper bound?

The same happens if you consider $$\int_0^1\frac{1}{\sqrt{x}}\,dx$$ with the substitution $$\sqrt{x}=t$$ or $$x=t^2$$ that transforms the integral into $$\int_0^1\frac{2t}{t}\,dt=\int_0^1 2\,dt=2$$ no longer an improper integral.

Note that an antiderivative of $$1/\sqrt{x}$$ is $$2\sqrt{x}$$, which is not differentiable at $$x$$.

• Not differential at $x=0$ you mean? – GambitSquared Nov 10 '18 at 8:46
• This is indeed a nice example, but I am trying to understand the principle of why substitution can eliminate the improperness of a integral. – GambitSquared Nov 10 '18 at 8:47
• @GambitSquared The “infinite” derivative cancels the singularity. – egreg Nov 10 '18 at 10:41

Rewrite your integral as $$\int^{t=\pi/2}_{t=0} \frac1{x(t)}\cdot\frac{dx(t)}{\sqrt{1-x(t)}}$$

Note that $$dx(t)=(\sin 2t)dt$$.

At $$t=\frac{\pi}2$$, there is a zero of order one in the $$\sqrt{1-x(t)}$$, however $$dx(t)$$ also has a zero of order one at $$t=\frac{\pi}2$$! The two zeroes exactly cancels out, leading to that $$\csc t$$ is not singular near $$t=\frac{\pi}{2}$$, as you observed.

You might feel uncomfortable to say ‘there is a zero in the differential’. Alternatively, rewriting your integral as $$\int^{t=\pi/2}_{t=0} \frac1{x(t)}\cdot\frac{\frac{dx(t)}{dt}}{\sqrt{1-x(t)}}dt$$, you may comfortably say that, at $$t=\frac{\pi}2$$, the first order zero in $$\sqrt{1-x(t)}$$ in the denominator is cancelled by the first order zero in $$\frac{dx(t)}{dt}$$.

Actually, we can always treat differentials as functions.

If you perform a contour integral of $$f(z)=\frac{z}{1+z^2}$$ along a small circle around $$z=\infty$$, i.e. $$-\oint_{|z|=\infty}\frac{z}{1+z^2}dz$$ one may expect the integral to be zero, since $$\frac z{1+z^2}$$ is analytic at $$z=\infty$$.

However $$dz$$ has a pole at $$z=\infty$$, thus indeed the integral is non-zero. It turns out that the integral equals $$2\pi i$$.