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

Answer 1

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

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

Answer 2

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:

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

Answer 3

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.

Answer 4

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,

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