Improper Integral

Topics: Calculus - Integral

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(definition)

Let $f:[a,\infty )\to \mathbb{R}$ an integrable function $\forall [a,t]\subseteq [a,\infty )$.

If ${\lim }_{t\to \infty }{\int }_a^{t}f$ exists, then we say that:

$${\int }_a^{\infty }f$$

…is the improper integral of $f$ in $[a,\infty )$ and that:

$${\int }_a^{\infty }f=\underset{t\to \infty }{\lim }{\int }_a^{t}fdt$$

Example

Let’s calculate ${\int }_1^{\infty }x{e}^{-x}dx$.

According to this definition, we have that:

$${\int }_1^{\infty }x{e}^{-x}dx=\underset{t\to \infty }{\lim }{\int }_1^{t}x{e}^{-x}dx$$

With the Fundamental Theorem of Calculus, we can write:

$${\int }_1^{\infty }x{e}^{-x}dx=\underset{t\to \infty }{\lim }\left[{-(x+1)e^{-x}|}_1^t\right]=\underset{t\to \infty }{\lim }\left(-\frac{t+1}{e^t}+\frac{2}{e}\right)$$

Then by evaluating the limit, we get the final result

$${\int }_1^{\infty }x{e}^{-x}dx=\frac{2}{e}$$

If such a limit doesn’t exist, then we say that such an integral diverges (i.e. it doesn’t converge).

Example of non-convergence

An example of an improper integral that diverges is ${\int }_0^{\infty }xdx$:

$${\int }_0^{\infty }xdx={\underset{t\to \infty }{\lim }\frac{x^2}{2}|}_0^t=\underset{t\to \infty }{\lim }\left(\frac{t^2}{2}-0\right)=\text{diverges}$$

Note that improper integrals don’t have to necessarily involve infinity. We can also have improper integrals when we calculate the integral of a function as it approaches a vertical asymptote.

Example of an improper integral that doesn't involve $\infty$

Let $f(x)=\frac{\sqrt{1+x}}{\sqrt{1-x}}$; $f(1)$ is indeterminate; $f$ has a vertical asymptote on $x=1$.

As such, ${\int }_0^1\frac{\sqrt{1+x}}{\sqrt{1-x}}$ is an improper integral:

= \lim_{t \to 1} \int_0^t \frac{\sqrt{1+x}}{\sqrt{1-x}}$$

= \lim_{t \to 1} \left[ \frac{\sqrt{1-t^2}}{2} + \arcsin \frac{\sqrt{1-t}}{\sqrt 2} - \left( \frac{1}{2} + \arcsin \frac{1}{\sqrt 2} \right) \right]

= -\frac{1}{2} - \frac{\pi}{6}