Why am I getting a finite integral for infinite area?

Question

The following is the graph of the $\tan(x)$ function:

We can clearly see there how it’s undefined at $\frac \pi 2$.

Now, what if we wanted to find the area between the tangent function and the x-axis in the interval $[0, 2]$?

The following happens:

$$\int \tan(x) dx = -\ln|\sec(x)| + c, \quad c\in\mathbb{R}$$

$$\implies \int_{0}^{2} \tan(x) = (-\ln|\sec(2)|) - (-\ln|\sec(0)|)$$

$$\implies (-\ln|\sec(2)|) - (-\ln(1)) = (-\ln|\sec(2)|) + \ln(1)$$

$$ |\sec(2)| = 2.40299796…$$

So what we end up getting is something in the form

$$\int_{0}^{2} \tan(x) dx = -\ln(2.40299..) + \ln(1)$$

Which ends up giving us

$$-0.87671710853 + 0$$

Which is definitely a finite value!

However, when you look at the above graph, you clearly see that there should be infinite area there, as the tangent is asymptotic one unit before there.

What’s going on here?

Answer 1

Your calculation has used the Fundamental Theorem of Calculus, but that theorem has an assumption that needs to be satisfied: the integrand must be continuous on the entire domain of integration. Yet $\tan x$ is not continuous on $[0,2]$.

This is a good reminder that theorems have hypotheses and we must always verify the hypotheses before applying the theorem.

Answer 2

Another thing is that infinite-looking areas can still have finite integrals.

For example, consider the function $1/x^2$ (start from $x=1$ but extending to infty):

Despite having infinite domain, the integral is finite:

$$ \displaystyle \int_{1}^{\infty} \dfrac{1}{x^2} dx = \left[ \dfrac{-1}{x} \right]_1^{\infty} = 1 $$

Answer 3

I disagree with Greg Martin that the central issue is that it's discontinuous. The central issue is with this claim: "$\int \tan(x) dx = -\ln|\sec(x)| $." By this, you presumably mean "For all $x$, $\frac {d}{dx} (-\ln|\sec(x)|)= \tan(x)$." However, the actual statement is "For all $x$ where $-\ln|\sec(x)|$ is defined, its derivative is $\tan(x)$." $\sec(x)$ isn't defined for $x=\frac{\pi}2+n\pi$, and its derivative doesn't exist there either, and so its derivative can't be equal to $\tan(x)$ (which also doesn't exist there). Now, the fact that it's not continuous there does indeed flow from the fact that it doesn't exist, so Greg Martin is correct in noting that it's not continuous and the Fundamental Theorem of Calculus doesn't apply, but the fact that the function doesn't exist is the primary issue.

When dealing with identities, especially with trigometric identities, you need to take care that you're taking into account for what values it applies.

Answer 4

In order to consider $$ \int_a^b f(x)\thinspace dx $$ you first of all need that the function $f$ is defined over $[a,b]$.

Now $f(x)=\tan x$ is not defined over $[0,2]$, because $\pi/2\in[0,2]$. OK, we can decide that $f(x)=\tan x$ for $x\in[0,2]$, $x\ne\pi/2$ and $f(\pi/2)=0$.

This might be done, but you cannot apply the fundamental theorem of calculus, which requires the integrand to be continuous over the interval the integration is carried on.

It is possible to extend the notion of integral to noncontinuous function, but this requires limits: you may interpret $$ \int_0^2 \tan x\thinspace dx=\lim_{a\to\pi/2^-}\int_0^a \tan x\thinspace dx+\lim_{b\to\pi/2^+}\int_b^2 \tan x\thinspace dx $$ provided both limits exist finite. Neither does: the first one is $\infty$, the second one is $-\infty$.

Answer 5

Multiple other people have answered why this integration isn't safe, but we do actually find that if we do it carefully it does work here.

However before we can do the integral we must first fix the error in the question. In the question you state that $$\int \tan(x) = -\log\left|\sec(x)\right|$$ but this is wrong and instead the integral is given by $$\int \tan(x) = \log\left|\sec(x)\right| = -\log\left|\cos(x)\right|$$

When you use the correct anti-derivative you get a positive area and in fact the correct area.

However we have to be careful to show that this is the correct area. While the function is discontinuous/not defined at $\pi/2$, it is well defined and continuous arbitrarily close to $\pi/2$ so we can define the integral to be the limit as we allow the range to be close and closer to including $\pi/2$.

As the discontinuity is in the middle of the range we will split the integral into two and take the limits to be $\delta$ away from $\pi/2$ which gives $$\int_0^2 \tan(x) = \lim_{\delta\to 0}\left(\int_0^{\pi/2-\delta} \tan(x) + \int_{\pi/2+\delta}^2 \tan(x)\right)$$ $$=\lim_{\delta\to 0}\left(\left.\log\left|\sec(x)\right|\right|_0^{\pi/2-\delta}+\left.\log\left|\sec(x)\right|\right|_{\pi/2+\delta}^2\right)$$ $$=\lim_{\delta\to 0}\left(\log\left|\sec\left(\pi/2-\delta\right)\right|-\log\left|\sec(0)\right|+\log\left|\sec(2)\right|-\log\left|\sec\left(\pi/2+\delta\right)\right|\right)$$ $$=\log\left|\sec(2)\right|-\log\left|\sec(0)\right|+\lim_{\delta\to 0}\left(\log\left|\sec\left(\pi/2-\delta\right)\right|-\log\left|\sec\left(\pi/2+\delta\right)\right|\right)$$ We will now use translation and reflection identities on the terms still in the limit to get $$=\log\left|\sec(2)\right|-\log\left|\sec(0)\right|+\lim_{\delta\to 0}\left(\log\left|\csc\left(\delta\right)\right|-\log\left|\csc\left(\delta\right)\right|\right)$$ Both terms in the limit are the same and as such cancel to give $$=\log\left|\sec(2)\right|-\log\left|\sec(0)\right|+\lim_{\delta\to 0}0$$ $$=\log\left|\sec(2)\right|-\log\left|\sec(0)\right|$$ Which is the same answer obtained by using the anti-derivative and ignoring the discontinuity.

Note however that while it works this time, some times the limit doesn't cancel and can either give a finite or infinite contribution.