Integral becomes improper after a substitution

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

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

• I've got a "wrong" feeling from this too. How was the substitution done? Since $\sin\frac\pi4=1/\sqrt2$ and $\sin\frac\pi2=1$, the substitution $u=\sin x$ would lead to $\int_{1/\sqrt2}^1\cdots\cdots(\cos u\,du)$. But you're not doing anything like $u=\sin x$. Th explanation is probably just that you did the substitution incorrectly. ${}\qquad{}$ – Michael Hardy Jun 24 '15 at 5:37
• You can see my substitution in my comment below. I will edit my initial question so everyone can see it. – asd Jun 24 '15 at 5:39
• @MichaelHardy by "below" I assume he means the comment under my answer. – Adam Hughes Jun 24 '15 at 5:45

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 ;)

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!

• Thank you for your answer. As I said, my transformation feels wrong and your integral looks perfectly right. It is maybe embarrassing, but I do not understand your notation for the substitution. It looks as if you use the 'physicists notation' which should give us the following: $z:=\sin(x)\Rightarrow dz=\sqrt{1-z^2}dx\Rightarrow dx=\frac{dz}{\sqrt{1-z^2}}$. Therefore we get the integral $\int \frac{dz}{z\cdot \sqrt{1-z^2}}$ (with appropriate domain). Can you please try to explain to me where my mistake is? – asd Jun 24 '15 at 5:35
• @asd again I believed your original substitution work too readily. I realized after reading your comment on Michael's answer: they should be arcsin limits, not sine limits. Check the modified version and you will see it agrees with the original at around $0.88$. – Adam Hughes Jun 24 '15 at 6:13
• @asd Indeed, note that $\arcsin(\pi/2)$ isn't even well-defined over $\Bbb R$, so the moral of the story is you made a bad substitution. – Adam Hughes Jun 24 '15 at 6:19
• @asd Glad to help you get things figured out. It reassures me as well to know that the technique of substitution is as rock solid as the theory says it ought to be! :-) – Adam Hughes Jun 24 '15 at 6:25
• @AdamHughes: I'm sorry, but I don't quite see how this answers the question. What specifically is wrong with $\int_{\pi/4}^{\pi/2}\frac{1}{\sin(x)}\,dx = \int_{1/\sqrt{2}}^1 \frac{1}{z\sqrt{1-z^2}}\,dz$? (Do you agree that these two integrals are equal?) The question is: why should the substitution $z = \sin(x)$, where $x \in (\pi/4, \pi/2)$ return an improper integral? – Jesse Madnick Jun 24 '15 at 9:55

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.

• No that is not my substitution. You have quoted the result of Adam Hughes. If we let mathematica calculate the integral $\int_{1/\sqrt{2}}^1 \cos(x)/\sin(\sin(x))$ it gives us the value approximately 0.28. whereas if we let mathematica calculate the initial integral it gives us 0.88. Therefore both integrals seems to be different – asd Jun 24 '15 at 5:51
• Yes, that's my computation of what he should get from his choice of substitution. He did it improperly, I was noting what it ought to be. – Adam Hughes Jun 24 '15 at 6:09
• @asd your limits are also wrong, again since your substitution was arcsin rather than sin, it should be from $\arcsin(\pi/4)$ to $\arcsin(\pi/2)$ which agrees with the original integral. – Adam Hughes Jun 24 '15 at 6:12

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:

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