Integral becomes improper after a substitution

Question

I'm suprised about the following phenomenon which I would like to discuss with you. Consider the proper integral $$\int_{\pi/4}^{\pi/2}\frac{1}{\sin(x)}dx.$$

Since $\sin(x)$ is a diffeomorphism on the interval $(\pi/4,\pi/2)$ we can make the substitution $\phi(x):=\arcsin(x)$. Since $d\phi(x)=\frac{dx}{\cos(\arcsin(x))}=\frac{dx}{\sqrt{1-x^2}}$ we get:

$$\int_{\pi/4}^{\pi/2}\frac{1}{\sin(x)}dx=\int_{1/\sqrt{2}}^{1}\frac{1}{x\cdot \sqrt{1-x^2}}dx.$$

Suddenly the integral becomes improper, which feels somehow 'wrong' to me. But I do not think that there is a mistake in the argumentation which leads from the initial integral to the improper, new one.

Do you know further examples similar to the above one? I hope someone of you can help me to eliminate my 'uncomfortable' or 'wrong' freeling about this situation, for -as the calculation shows- the situation is quite natural.

Best wishes

edit: Another (maybe more familiar) notation to write the substitution would be:

$z:=\sin(x)\Rightarrow dz=\cos(x)dx=\sqrt{1-z^2}dx\Rightarrow \frac{dz}{\sqrt{1-z^2}}=dx$. Hence we get the integral

$$\int_{1/\sqrt{2}}^{1}\frac{1}{z\cdot\sqrt{1-z^2}}dz$$

Answer 1

To answer the original question, which is psychological more than mathematical, it is perfectly reasonable to intuit that a proper integral should remain proper under substitution. After all, the area does not change.

The source of the impropriety comes from $\sin(x)$ having zero derivative at $x = \frac{\pi}{2}$. Because its value changes very slowly in the vicinity of $\frac{\pi}{2}$, there is a lot of "area" compressed into an infinitesimal change in $\sin(x)$. This is not a concern when you are integrating w.r.t. $x$, but it is when you are integrating w.r.t. $\sin(x)$. To compensate, the value of the integrand must tend to infinity at this point, which results in an improper integral.

Or, consider that $dx = \frac{dx}{dz}dz$. As $\frac{dz}{dx} \rightarrow 0$, $\frac{dx}{dz} \rightarrow \infty $.

Or, consider the slightly simpler case of $$\int_0^1 \left(1-x^2\right) dx$$

Doing a similar substitution $ y := x^2 \implies \frac{dy}{2\sqrt y} = dx$, you get another improper integral:

$$\int_0^1 \frac{1-y}{2\sqrt{y}}dy$$

...and it is again because $\frac{d}{dx}x^2 = 0$ at $x = 0$.

Also, in "$\phi(x):=\arcsin(x)$", asd obviously meant for $\phi$ to be $x$, and $x$ to be the new variable, which is why he later called it $z$ for clarity. The intended meaning was a perfectly valid substitution, as Mathematica confirmed ;)

Answer 2

Indeed $d\phi ={dx\over\sqrt{1-x^2}}$, however you do not have a $d\phi$ in your integral, you have $dx=\cos\phi\,d\phi$, so what you are looking at is

$$\int_{\arcsin(\pi/4)}^{\arcsin(\pi/2)}{\cos x\over \sin(\sin(x))}\,dx$$

which is perfectly proper, at least at a glance. But note that what you really did wrong was you ($1$) did the substitution wrong and ($2$) picked a bad function: $\arcsin$ is only defined on $[-1,1]$ and you have used it all the way up to $\pi/2>1$, so it's not really even a valid choice!

Answer 3

Your comment under Adam Hughes' answer says you've got $$ \int_{\sqrt{2}^{-1}}^1 {\cos\phi\,d\phi \over \sin(\sin\phi)}. $$ There's nothing improper about that.

Answer 4

The change of variable that you chose induces the "improperness", as the derivative of $\sin(x)$ vanishes at $x=\pi/2$, so that $dz=\cos(x)\,dx$ degenerates.

You have a similar phenomenon with the substitution $x=e^t$ in the integral below:

$$\int_0^1\,dx=\int_{-\infty}^0 e^{t}\,dt=e^t\Big|_{-\infty}^0.$$