Fundamental Theorem of Calculus Confusion regarding atan

According to this site,

$$\int \frac{1}{a^2 \cos^2(x) + b^2 \sin^2(x)} \,dx =\frac{1}{ab} \arctan\left(\frac{b}{a} \tan(x)\right)$$

Thus, $$\int_0^{\pi} \frac{1}{a^2 \cos^2(x) + b^2 \sin^2(x)} \, dx =\frac{1}{ab} \arctan\left(\frac{b}{a} \tan(\pi)\right)-\frac{1}{ab} \arctan\left(\frac{b}{a} \tan(0)\right)=0-0=0$$

But in fact value of integral is not zero.

What am I doing wrong?

• My first thought is that the arctangent function is somewhat problematic. It is an inverse only of a restriction of the tangent function. $\qquad$ – Michael Hardy Mar 14 '16 at 17:39
• I believe it is because $\arctan(0)$ is not always $0$, e.g. it could also be $\pi$. In that sense $\arctan$ is a multi-valued function. – Mufasa Mar 14 '16 at 17:39
• Let $a=b=1$. Then $\frac{1}{ab} \arctan\left(\frac{b}{a} \tan(\pi)\right) =\frac{1}{ab} \arctan 0 = 0 \ne \pi$. Your equation only works on $(-\pi, \pi)$ – steven gregory Mar 14 '16 at 17:46
• See this graph. Your function is not continuous, that's why it works only on $(-\pi/2;\pi/2)$ – Kamil Jarosz Mar 14 '16 at 17:58
• Some related questions: math.stackexchange.com/questions/1296351/…, math.stackexchange.com/questions/1356523/… – Hans Lundmark Mar 14 '16 at 18:29

First, let us assume that $a, b > 0$ without loss of generality. Then in this regime, $f(x) = (a^2 \cos^2 x + b^2 \sin^2 x)^{-1}$ is clearly a well-defined function everywhere, and is periodic with period $\pi$. However, the antiderivative is not unique, and the particular choice of antiderivative $$F(x) = \frac{1}{ab} \tan^{-1} \left( \frac{b}{a} \tan x \right)$$ may not be continuous over a given interval. For example, if we choose the branch of $\tan^{-1}$ that restricts the range to lie in $(-\pi/2, \pi/2)$, then you get this picture for $a = 2$, $b = 0.5$:

As you can see, there is a jump discontinuity in the antiderivative at $x = \pi/2$ that could be rectified by choosing a different branch of the inverse tangent; e.g., we could choose $\tan^{-1}$ to have the range $[0,\pi)$. This fixes the discontinuity at $\pi/2$ so that a definite integral containing this point will not result in an incorrect evaluation.

In general, it is important when dealing with branches of multivalued functions to choose the branch appropriate to the interval or domain of integration, so as to avoid these problems.

An example of how you could specify an antiderivative that is continuous for all real $x$ provided that we state that the range of $\tan^{-1}$ is $(-\pi/2,\pi/2)$, is by writing it as $$F(x) = \begin{cases} \displaystyle \frac{1}{ab} \left( \tan^{-1} \left(\frac{b}{a} \tan x \right) + \pi \left(\lfloor 2x/\pi \rfloor - \lfloor x/\pi \rfloor\right) \right), & 2x/\pi \not \in \{\pm 1, \pm 3, \ldots\} \\ \displaystyle \frac{x}{ab}, & 2x/\pi \in \{\pm 1, \pm 3, \ldots\}. \end{cases}$$ But I think this is a rather complicated way to express the antiderivative. Instead, it might be better to state that the branch of the inverse tangent that is chosen depends on the argument of $x$; i.e., if we define $y = \tan^{-1} c \tan x$ for some $0 < c < \infty$, then $y$ is chosen such that $x, y \in ((2m-1)\pi/2, (2m+1)\pi/2)$ for a suitable integer $m$.

• What is the value of antiderivative then? Where should a shift occur? – Andrew Fount Mar 14 '16 at 19:43
• So, what is antiderivative in this particular case? If we take integer m then 2m-1 is never 0. How we then choose interval (0,pi)? – Andrew Fount Mar 14 '16 at 21:03
• You've completely misunderstood the issue. When I wrote an alternative form of $F(x)$, that form uses a definition of inverse tangent such that $\tan^{-1} x \in (-\pi/2, \pi/2)$. Plot it yourself and you will see it is continuous. If you choose to define $\tan^{-1} x$ in a way that its range is on $[0,\pi)$, then that formula doesn't work. The formula doesn't determine the choice of branch: it is your choice of branch that determines what a formula looks like. – heropup Mar 14 '16 at 21:18
• That's the question. What is the formula for inverse tangent, whose range is [0, pi)? – Andrew Fount Mar 15 '16 at 12:19

Note that, in the link, $u=\tan x$ which is increasing in $[0,\pi/2)$. In this integral, $x\in[0,\pi]$ and hence you can't use the result directly. In fact

$$\int_0^\pi \frac{1}{a^2 \cos^2(x) + b^2 \sin^2(x)} \,dx=2\int_0^{\pi/2} \frac{1}{a^2 \cos^2(x) + b^2 \sin^2(x)} \,dx.$$ Use the result in $[0,\pi/2]$ and you will get the result.

• So, my result is wrong because tan is infinite at pi/2? – Andrew Fount Mar 14 '16 at 17:47
• Recall that in your result you will actually have a term like this:$$\frac{1}{ab}\arctan(\frac{b}{a}\tan(\frac{\pi}{2}))=\frac{1}{ab}\arctan( \infty )=\frac{1}{ab}\frac{\pi}{2}$$ – Mufasa Mar 14 '16 at 17:51
• @AndrewFount, because $y=\tan x$ is not increasing in $[0,\pi]$. – xpaul Mar 14 '16 at 17:57
• @Mufasa, you are right. – xpaul Mar 14 '16 at 17:58
• @Andrew Fount -- that is essentially correct. The tan function goes to infnity and then jumps to - infinity and that introduces a discontinuity. – Doug M Mar 14 '16 at 18:08

Some Background on Integration by Substitution

It is important to understand the assumptions one makes in Integration by Substitution.

Let $I$ denote the integral given by $I=\int_a^b f(x)\,dx$.

Condition $(i)$: We assume that $f$ is continuous on $[a,b]$.

Condition $(ii)$: The inverse function $u^{-1}$, of $u(x)$ is continuously differentiable on $[u(a),u(b)]$.

Under Conditions $1$ and $2$, we have

\begin{align}I=&\int_a^b f(x)\,dx\\\\&=\int_{u(a)}^{u(b)}f(u^{-1}(t))\frac{du^{-1}(t)}{dt}\,dt \tag 1\end{align}

Note that the relationship expressed in $(1)$ can be obtained formally through the substitution $u(x)=t$, under Assumptions $(i)$ and $(ii)$.

Application to Evaluating $\int_0^\pi \frac{1}{a^2\sin^2(x)+b^2\cos^2(x)}\,dx$

To apply the substitution in $(1)$ correctly to evaluate the integral of interest, one must ensure that both $(i)$ and $(ii)$ are satisfied.

Since $f(x)=\frac{1}{a^2\sin^2(x)+b^2\cos^2(x)}$ is continuous for $x\in [a,b]$ and Condition $1$ is satisfied.

The substitution used in the reference cited in the OP is $\tan(x)=t$. Note that for $x\in [0,\pi]$ the inverse function for the tangent function is given by

$$\tan^{-1}(t)=\begin{cases}\arctan(t)&, t>0\\\\\pi+\arctan(t)&,t<0 \tag 2\end{cases}$$

where the $-\pi/2 < \arctan(x)<\pi/2$ is the principal branch of the inverse tangent function. The inverse tangent is not even continuous at $0$ and we cannot apply, therefore, the substitution over the interval $x\in [0,\pi]$.

Instead, we can write the integral as the sum

\begin{align} \int_0^\pi f(x)\,dx&=\lim_{\epsilon \to 0^+}\left(\int_0^{\pi/2-\epsilon}f(x)\,dx+\int_{\pi/2+\epsilon}^\pi f(x)\,dx\right) \end{align}

where $f(x)=\frac{1}{a^2\sin^2(x)+b^2\cos^2(x)}$.

The inverse tangent function $\tan^{-1}(x)$ as expressed in $(2)$ is continuously differentiable on each separate interval $x>0$ and $x<0$ and so, Condition $2$ is satisfied on each separate interval.

Then, we can write

\begin{align}\lim_{\epsilon \to 0^+}\int_0^{\pi/2-\epsilon}f(x)\,dx&=\int_0^{\infty}f(\arctan(t))\frac{1}{1+t^2}\,dt\\\\ \lim_{\epsilon \to 0^+}\int_{\pi/2+\epsilon}^\pi f(x)\,dx&=\int_{-\infty}^0f(\pi+\arctan(t))\frac{1}{1+t^2}\,dt \tag 3 \end{align}

Since $f(x)$ is even and $\pi$-periodic, then the second equality in $(3)$ is identical to the first. Therefore, we have

\begin{align} \int_0^\pi f(x)\,dx&=2\int_0^\infty \frac{f(\arctan(t))}{1+t^2}\,dt\\\\ &=2\int_0^\infty \frac{1}{a^2t^2+b^2}\,dt\\\\ &=\left. \frac{2\arctan(at/b)}{ab}\right|_{0}^{\pi/2}\\\\ &=\frac{\pi}{ab} \end{align}

This very integral, with a little manipulation, is referenced in the Jeffrey paper so the nice form in that paper can be attained by observing that $\tan^{-1}(\tan x)$ has the same discontinuities as the original integral, so we may hope to cancel them out by setting $$\frac1{ab}\tan^{-1}\left(\frac{b}a\tan x\right)=\frac1{ab}\left\{x+\tan^{-1}\left(\frac{b}a\tan x\right)-\tan^{-1}(\tan x)\right\}$$ Now, \begin{align}\tan^{-1}\left(\frac{b}a\tan x\right)-\tan^{-1}(\tan x) & =\tan^{-1}\left(\frac{\frac{b}a\tan x-\tan x}{1+\frac{b}a\tan^2x}\right) \\ & =\tan^{-1}\left(\frac{(b-a)\sin2x}{(a+b)-(b-a)\cos2x}\right) \end{align} after a little algebra, so $$\int\frac1{a^2\cos^2x+b^2\sin^2x}dx=\frac1{ab}\left\{x+\tan^{-1}\left(\frac{(b-a)\sin2x}{(a+b)-(b-a)\cos2x}\right)\right\}+C$$ removes all discontinuities and even checks by differentiation.