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Why is it legal to change the variable of an indefinite integral?


Consider $$\int \dfrac{dx}{\cos x}$$

If one were to say, $\text{Let } u=\cos x$, do we not now technically have $$\int_{u=\cos-\infty}^{u=\cos\infty}u^{-1}\dfrac{d\arccos u}{du} \cdot du$$ Which is clearly madness even before we consider that whatever $\cos \infty$ is, $\cos -\infty$ is too.


So my question is simply, what have I missed; why is it allowed?

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  • $\begingroup$ If you consider $u=\cos x$ then $x=\arccos u.$ So $dx=\frac{-1}{\sqrt{1-u^2}}du.$ Where is $\arccos u$ in the integral? On the other hand, why you try to convert and indefinite integral into a definite integral? $\endgroup$ – mfl May 13 '14 at 19:21
  • $\begingroup$ Woops. That was a dumb error. Well an indefinite is no different to 'definite over infinite range in each direction' right? $\endgroup$ – OJFord May 13 '14 at 19:28
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    $\begingroup$ They are completely different things. An indefinite integral is a function (if we assume some normalization on the constant of integration) and a definite integral is a number (if it exits). $\endgroup$ – mfl May 13 '14 at 19:31
  • $\begingroup$ But am I wrong in saying $\int_{-\infty}^{\infty}y(x)dx := \int y(x)dx$? $\endgroup$ – OJFord May 13 '14 at 19:39
  • $\begingroup$ I'm not "trying to turn it into a definite integral" - I'm just asking why it is allowed to change the variable. If it was definite, the values of those bounds would also change. $\endgroup$ – OJFord May 13 '14 at 19:41
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You don't even hint at the nature of the mysterious process that led you to the conclusion that the bounds of integration would be $\pm\infty$. Let's try it with some actual bounds: $$ \int_0^{\pi/4}\frac{dx}{\cos x} = \int_1^{\sqrt{2}/2} \frac{d\arccos u}{u} = \int_1^{\sqrt{2}/2} \frac{-du}{u\sqrt{1-u^2}} $$ No "madness" is involved.

If you want to think about $\displaystyle\int_{-\infty}^\infty \frac{dx}{\cos x}$, you'll meet some unpleasant complications. $$ \int_{-\infty}^\infty \frac{dx}{\cos x} = \sum_{n=-\infty}^\infty \int_{2\pi n}^{2\pi(n+1)} \frac{dx}{\cos x} = \sum_{n=-\infty}^\infty \int_0^{2\pi} \frac{dx}{\cos x}. $$ That last sum would be $0$ if the value of the integral is $0$ and either $\pm\infty$ if it is any other number. So let's think about $$ \int_0^{2\pi} \frac{dx}{\cos x} = \left(\int_{-\pi/2}^{\pi/2} + \int_{\pi/2}^{3\pi/2} \right) \frac{dx}{\cos x} = \infty + (-\infty). $$ Perhaps one could say that a sort of Cauchy principal value of this integral is $0$. At each of the two locations $a$ of vertical asyptotes, one would have $$ \lim_{\varepsilon\downarrow0}\left(\int_\cdots ^{a-\varepsilon} + \int_{a+\varepsilon}^ \cdots\right). $$ Then one would get $0$.

But one should avoid substitutions that are not one-to-one.

As to why substitutions are valid, the short answer, which you'll find in every textbook is: the chain rule.

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  • $\begingroup$ When I was first taught to integrate, I was told $\int = \int_{-\infty}^{\infty}$. I never really thought too much about it again until now. $\endgroup$ – OJFord May 13 '14 at 20:41
  • $\begingroup$ I never said there was anything mad about the definite integral, just misunderstood the indefinite. $\endgroup$ – OJFord May 13 '14 at 20:42
  • $\begingroup$ Further, I clearly understood that the chain rule allows substitutions of definite integrals, since I applied it in my OP. There's really no need for such condescension.. $\endgroup$ – OJFord May 13 '14 at 20:44
  • $\begingroup$ What you say you were told I've never seen before. $\endgroup$ – Michael Hardy May 13 '14 at 20:45
  • $\begingroup$ Okay, sorry I misunderstood.. It was a while ago, I think we were taught definite first (as the application to area under curve was emphasised far more than proof), so from there I guess we were told indefinite "is like definite, but from $-\infty$ to $\infty$" or such. I don't know. $\endgroup$ – OJFord May 13 '14 at 20:52

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