Why we get area under curve when function goes infinity between limits?

Question

Why we get area under curve of definite integral between limits where function goes to infinity using substitution method.

e.g : $$ \int_{0}^{\pi} \sec(x).\tan(x) \space dx $$ Function goes to infinity between these limits.why we get area under curve?

Answer 1

I will try to give some background here, hope this will help the OP to understand what is going on with this curve and this integral.

First, "area under curve" is a vague concept until defined rigorously. Let's not get into the intricacies of Jordan measure and Lebesgue measure, but if the curve is defined by a non-negative function $f:[a,b]\to\mathbb R$, then it turns out that:

• If $f$ is Riemann-integrable on $[a,b]$ (which implies it is bounded), then the Jordan measure of "area under curve" exists and is equal to $\int_a^b f(x)dx$ (Riemann integral)
• If $f$ is Lebesgue-integrable on $[a,b]$, then the Lebesgue measure of "area under curve" exists (even though it can be infinite), and it is equal to $\int_a^b f(x)dx$ (Lebesgue integral)

As it happens, Riemann integral is simpler to teach (and is so usually introduced first to students), but is more restricted in terms of when it can be calculated. For one thing, it requires the function to be bounded, which your function certainly isn't. (Your function is $f(x)=\sec x\cdot\tan x=\frac{\sin x }{\cos^2 x}$ and, as you've noticed, $\lim_{x\to\pi/2}f(x)dx=+\infty$). You can still use Lebesgue integral, or you can use the sum of the improper Riemann (or Lebesgue) integrals: $\int_0^{\pi/2-0}f(x)dx+\int_{\pi/2+0}^\pi f(x)dx$. Both approaches can be proven to give you the same result.

Now, as $\frac{1}{\cos x}$ is the primitive function of your function, it follows that $\int_a^b \frac{\sin x}{\cos^2 x}dx=\frac{1}{\cos x}\mid_a^b=\frac{1}{\cos b}-\frac{1}{\cos a}$ on any interval where your function is defined and Riemann-integrable (Fundamental theorem of Calculus). This precludes intervals such as $[0,\pi]$, but still lets us calculate the improper integrals:

$$\int_0^{\pi/2-0}f(x)dx=\lim_{b\to\pi/2-0}\left(\frac{1}{\cos b}-\frac{1}{\cos 0}\right)=+\infty$$ $$\int_{\pi/2+0}^\pi f(x)dx=\lim_{a\to\pi/2+0}\left(\frac{1}{\cos \pi}-\frac{1}{\cos a}\right)=+\infty$$

and so, by adding up those two numbers, the final integral diverges, i.e. its value is $+\infty$, which also means that the area under the curve is infinite.

(At least, this happens with the function from your question... You can think of another function, say $f(x)=\frac{\sin x}{\sqrt{|\cos x|}}$, for which the improper integrals would be finite and their sum would give you a finite "area under curve" - give it a try!)

Note a common mistake one can make: to calculate

$$\int_0^\pi f(x)dx=\frac{1}{\cos x}]_0^\pi=\frac{1}{\cos \pi}-\frac{1}{\cos 0} = 1-1 = 0$$

This is obviously not just incorrect but doesn't make any sense (as the curve is obviously "above" the $x$-axis, the area "under" it cannot end up being zero!), but as pointed above the mistake is that one has used the Fundamental theorem of Calculus on an interval where that theorem does not apply.

Answer 2

By symmetry (i.e. by applying the substitution $x\mapsto \pi-x$ on the integration range $(\pi/2,\pi]$) $$\int_{0}^{\pi}\frac{\sin(x)}{\cos^2(x)}\,dx = 2\int_{0}^{\pi/2}\frac{\sin(x)}{\cos^2(x)}\,dx=2\left[\frac{1}{\cos x}\right]_{0}^{\pi/2} $$ is divergent. On the other hand $$ \int_{0}^{\pi}\frac{\sin(x)}{\left[\cos^2(x)\right]^a}\,dx = \frac{2}{1-2a} $$ for any $a<\frac{1}{2}$.

A function doesn't need to be bounded to be improperly Riemann integrable. Think to $\frac{1}{\sqrt{x}}$ over $(0,1]$, for instance.

Answer 3

See Math HWS: Improper Integrals with Infinite Discontinuities. Ordinarily, when $f$ has an infinite discontinuity at $a$ or $b$, the integral $\displaystyle \int_a^b f(x)\,\mathrm{d}x$ is defined with the limit as the bounds approach $a$ or $b$, e.g., $\displaystyle \int_a^{b-\varepsilon} f(x)\,\mathrm{d}x$ as $\varepsilon\to0$. Then when there is an infinite discontinuity at $c\in(a,b)$, you'd split the integral into $\displaystyle \int_a^c f(x)\,\mathrm{d}x+\int_c^b f(x)\,\mathrm{d}x$ and treat each integral with the previous rule. For example, in the case of $\displaystyle f(x)=-\ln(x^2)$ on $[-1,1]$, which has a discontinuity at $0$ that rapidly approaches $+\infty$, you would have

$$\int_{-1}^1 f(x)\,\mathrm{d}x=\lim_{\varepsilon \to 0} \left[ \int_{-1}^{0-\varepsilon} -\ln(x^2)\,\mathrm{d}x +\int_{0+\varepsilon}^{1} -\ln(x^2)\,\mathrm{d}x \right] \\ =\lim \left(\left[-x (\ln(x^2)-2)\right]_{-1}^{-\varepsilon} +\left[-x (\ln(x^2)-2)\right]_{\varepsilon}^1\right) \\ =\lim \left(2\varepsilon\left(\ln\left(\varepsilon^{2}\right)-2\right)+4\right) \\ =4$$

Since the functions inside the limit are being integrated over continuous intervals and the limit exists, the integral is finite despite the discontinuity. However, if we tried this in the case of $f(x)=\sec x\tan x$ we would see that the limit does not exist since the function's discontinuity doesn't approach $+\infty$ quickly enough.