I recently came across the integral:

$$ \int \sqrt{(x-3)}\Big[\arcsin(\ln x) + \arccos(\ln x)\Big]dx. $$

It is easy to compute the integral without observing domain of integrand and answer comes out as $$ \frac{\pi}{3(x-3)^{(3/2)}} + C $$ using $\arcsin(t) + \arccos(t)=\pi/2$.

So is it allowed to integrate like this or can I conclude that the function does not have any indefinite integral. Kindly help.

  • 1
    $\begingroup$ How are you defining integration with a function that has no domain? $\endgroup$ – Peter Foreman Apr 23 '20 at 10:14
  • $\begingroup$ usually for complex integrals especially in indefinate integrals we usually do not consider domain of integrand and apply some substitutions or properties which is what i did here. $\endgroup$ – Ginger bread Apr 23 '20 at 10:34
  • $\begingroup$ I don't know who "we" refers to but it is not possible for any integral, definite or otherwise, to be performed without knowing that the integrand has a non-empty domain. $\endgroup$ – Peter Foreman Apr 23 '20 at 14:02

If $f$ is defined at no point, then $\int f(x)\;dx$ cannot be computed. But:

If $f$ has complex values, such as $$ f(x) := \sqrt{(x-3)}\Big[\arcsin(\ln x) + \arccos(\ln x)\Big] $$ then $\int f(x)\;dx$ can often be computed. Of course you have to verify that $\arcsin(\ln x) + \arccos(\ln x) = \pi/2$ even for cases where the $\arcsin$ and $\arccos$ are non-real (by looking up the definitions in that case).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.