Q.The value of $\displaystyle\int_0^{2/3}\dfrac{dx}{\sqrt{4-9x^2}}$ is

A Riemann (definite) integral $\int_a^b f(x)\,dx$ requires a finite interval $[a,b]$ with $f$ finite throughout. An improper integral of the first kind relaxes the finiteness of the interval — one or both limits of integration are $\pm\infty$ — and is defined as a limit of ordinary Riemann integrals:

If the limit exists (and is finite), the improper integral is said to converge; otherwise it diverges. By the Second Fundamental Theorem, if $F$ is an anti-derivative of $f$, then $\int_a^\infty f(x)\,dx=\lim_{t\to\infty}\big[F(t)-F(a)\big]$ — so in practice one finds $F$, substitutes $t$ for the infinite endpoint, and takes the limit termwise.

Recurring limiting facts used throughout this chapter:

• $\displaystyle\lim_{t\to\infty}\tan^{-1}t=\dfrac{\pi}{2}$, so $\int_0^\infty \dfrac{dx}{a^2+x^2}=\dfrac{\pi}{2a}$ for $a>0$.
• $\displaystyle\lim_{x\to\infty}e^{-x}=0$ and, by L'Hôpital's rule applied repeatedly, $\displaystyle\lim_{x\to\infty}x^m e^{-x}=0$ for every positive integer $m$ — a polynomial can never outrun $e^{-x}$'s decay, which is exactly what makes $\int_0^\infty x^n e^{-x}\,dx$ (the Gamma integral) converge.

A convenient substitution often converts a proper-looking trig integral on $[0,\pi/2]$ into an improper integral in disguise: putting $u=\tan x$ sends $x=\pi/2$ to $u=\infty$, turning $\int_0^{\pi/2}\dfrac{\sec^2x}{4\tan^2x+5}\,dx$ into the improper integral $\int_0^\infty\dfrac{du}{4u^2+5}$, evaluated exactly the same way.