Mistake in definite integral of $\int_{-1}^{1} d/dx \arctan(1/x) dx$

Question

$$\int_{-1}^{1} \left(\frac{\mathrm{d}}{\mathrm{d}x}\arctan{\frac1x}\right) \, \mathrm{d}x$$

Since integration cancels the differentiation, we need $\arctan{(1)} - \arctan{(-1)}$, that is $\pi/2$

But in the book the answer is $-\pi/2$ , so I think its a mistake

Books are right 99.99% of the time, and we are not in the remaining 0.01% of the cases here. — , Mar 30 '18 at 13:06

King Tut · Answer 1 · 2018-03-30T13:19:35.010

For a simple check, just differentiate it and try to integrate. That should give you a satisfactory answer:

$$\int_{-1}^{1} \frac{\mathrm d}{\mathrm dx}\arctan\left(\tfrac{1}{x}\right)\,\mathrm dx = -\int_{-1}^{1} \frac{1}{1+x^2}\,\mathrm dx = -\frac{\pi}{2}$$

I think issue is as pointed in comments, the function $\arctan(\tfrac{1}{x})$ is not continuous over the chosen interval on integral, discontinuity at $x=0$. So you cannot apply fundamental theorem of calculus, but if you split the integral, then you can apply the theorem:

$$\int_{-1}^{1} \frac{\mathrm d}{\mathrm dx}\arctan\left(\tfrac{1}{x}\right)\,\mathrm dx = \int_{-1}^{0} \frac{\mathrm d}{\mathrm dx}\arctan\left(\tfrac{1}{x}\right)\,\mathrm dx + \int_{0}^{1} \frac{\mathrm d}{\mathrm dx}\arctan\left(\tfrac{1}{x}\right)\,\mathrm dx $$

Andrew Li I was editing, i lost all text!! :( But thanks! :) — King Tut, Mar 30 '18 at 13:12
It shouldn't have kicked you out... it only does when you click 'reload' right? — Andrew Li, Mar 30 '18 at 13:14

user · Accepted Answer · 2018-03-30T13:24:51.843

0

To avoid the problem for $x=0$, note that the following holds

for $x>0 \implies \arctan \frac 1 x=\frac{\pi}2 -\arctan x $
for $x<0 \implies \arctan \frac 1 x=-\frac{\pi}2 -\arctan x $

then

$$\int_{-1}^{1} \frac{d}{dx}\arctan(1/x) dx =\int_{0}^{1} \frac{d}{dx}\arctan(1/x) dx +\int_{-1}^{0} \frac{d}{dx}\arctan(1/x) dx \\=-\int_{0}^{1} \frac{d}{dx}\arctan(x) dx -\int_{-1}^{0} \frac{d}{dx}\arctan(x) dx $$

edited Mar 30 '18 at 13:24

answered Mar 30 '18 at 13:08

user

154,566

Hmm nice one, is this a property? – King Tut Mar 30 '18 at 13:15
@KingTut Yes of course, even if I should precise it better – user Mar 30 '18 at 13:21
Exactly the way to get the wrong answer because this removes the discontinuity. – Mar 30 '18 at 13:21
@Yves I dont understand, does this not give $\pi - \pi/4 - (\pi+\pi/4) = -\pi/2$ ? – King Tut Mar 30 '18 at 13:22
@KingTut Now it should be ok, there was indeed a problem in the previuos version! – user Mar 30 '18 at 13:25
@YvesDaoust Yes indeed the idea is to eliminate the discontinuity at $x=0$ – user Mar 30 '18 at 13:27

score 0 · Answer 3 · answered Mar 30 '18 at 13:11

0

Just observe that

$$\int_{-1}^1 d[\arctan(1/x)]=\arctan(1)-\lim_{x\to0^+}\arctan(1/x)+\lim_{x\to 0^-}\arctan(1/x)-\arctan(-1)$$

That is: the above is an improper integral.

answered Mar 30 '18 at 13:11

Masacroso

30,417

score 0 · Answer 4 · answered Mar 30 '18 at 13:16

The integrand is undefined at $x=0$ because of the discontinuity of the $\text{arccot}$ function. Hence the fundamental theorem of calculus cannot be applied blindly.

If you split the integration domain,

$$\int_{-1}^{0^-}d\text{ arccot }x+\int_{0^+}^1d\text{ arccot }x=\text{ arccot }0^-+\text{ arccot }1+\text{ arccot }1-\text{ arccot }0^+\\ =-\frac\pi2+\frac\pi4+\frac\pi4-\frac\pi2.$$

Mistake in definite integral of $\int_{-1}^{1} d/dx \arctan(1/x) dx$

4 Answers4