Tricky integral question

Question

Find the value of : $I=\int\limits_{-1}^1 \left[ \frac{d}{dx}( \tan^{-1} \left( \frac{1}{x} \right) \right]\ dx$

My approach: $I=\int_{-1}^{1}d(\tan^{-1}(\frac{1}{x})) = (\tan^{-1}(1/x))^{1}_{-1} = \tan^{-1}(1)-\tan^{-1}(-1) = \frac{π}{4}-(-\frac{π}{4})=\frac{π}{2}$.

Solution given: $$\frac{d}{dx}(\tan^{-1}(\frac{1}{x})) = \frac{d}{dx}(\cot^{-1}(x)) = \frac{-1}{1+x^2}$$

$$\implies I= -(\tan^{-1}(x))^{1}_{-1} = -\frac{π}{2}$$

Why is my approach wrong?

Answer 1

Need to split the interval in $[-1,0];[0,1]$

Method $\#1:$

Use Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?

For $x>0,$

$\arctan\dfrac1x=$arccot$(x)=\dfrac\pi2-\arctan(x)$

For $x<0,$

$\arctan\dfrac1x=-\arctan\left(-\dfrac1x\right)=-$arccot$(-x)=-(\pi-$arccot$x)=-\pi+\dfrac\pi2-\arctan(x)$

Method$\#2:$

Alternatively using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

$\arctan x+\arctan\dfrac1x=$sign$(x)\cdot\dfrac\pi2$

Answer 2

This is almost a comment, rather than an answer, but as the comments above were almost answers, it seems fair.

To given another example, which might make clear what is going wrong:

Suppose $$ f(x) = \cases{-1, & x<0,\\ 0, & x=0,\\ 1, & 0 <x.}$$

Now, except at $x=0$, $f'(x) =0$. (The graph of $f$ consists basically of two lines of slope $0$.)

Suppose that the question were to evaluate $$I= \int_{-1}^1{d\over dx}f(x) \,dx.$$

The 'FTC method' would give $I= f(1) - f(-1) = 1- (-1) = 2$.

On the other hand, as $f'(x) = 0$ (except at $x=0$!), one should certainly get (geometrically, say, or Riemann sums) that $$ \int_{-1}^1{d\over dx}f(x) \,dx \,= \int_{-1}^1 0\,dx = 0.$$ This is in fact the 'correct' answer.

This is analogous to the question you were given.

I think that one can object that the question used the expression, ${d\over dx} f(x)$, which is only defined 'almost everywhere.' In the framework of a calculus class and Riemann integration, $I$ should be treated as an improper integral - and the given answer [although 'correct'!] does not do so... If one treats the integral as an improper integral (i.e., break up the integral around $x=0$, and take limits), and use your 'FTC' method, one gets the correct answer.

Likewise, the given solution uses $$\frac{d}{dx}(\tan^{-1}(\frac{1}{x})) = \frac{d}{dx}(\cot^{-1}(x)) = \frac{-1}{1+x^2}.$$ But that equality fails at $x=0$.

FWIW, there is a 'theory of integration' (Lebesgue) that handles 'almost everywhere,' and does not make a distinction between proper and improper integrals.

Answer 3

Here we will address your integral: \begin{equation} \int_{-1}^{1} \frac{d}{dx} \left[\arctan\left(\frac{1}{x}\right) \right]\:dx \end{equation} Now: \begin{equation} \frac{d}{dx} \left[\arctan\left(\frac{1}{x}\right) \right] = \frac{1}{1 + \left(\frac{1}{x}\right)^2} \cdot -\frac{1}{x^2} = -\frac{1}{1 + x^2} \end{equation} Thus your integral becomes: \begin{align} \int_{-1}^{1} \frac{d}{dx} \left[\arctan\left(\frac{1}{x}\right) \right]\:dx &= \int_{-1}^{1} -\frac{1}{1 + x^2}\:dx = -2\int_0^1 \frac{1}{1 + x^2}\:dx \nonumber \\ &= -2\left[\arctan(x) \right]_0^1 = -2 \cdot \frac{\pi}{4} = -\frac{\pi}{2} \end{align}

Answer 4

There is a much easier way of doing this, note that: $$\int_{-1}^1\frac{d}{dx}\left[\arctan\left(\frac 1x\right)\right]dx=\int_{-1}^1d\left(\arctan\left(\frac 1x\right)\right)=\left[\arctan\left(\frac 1x\right)\right]_{-1}^1$$