Question

Express each improper integral in Exercises15–20 as a sum of limits of proper definite integrals. Do not calculate any integrals or limits; just write them down.

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

Short answer

The integral is,

$\underset{x\to -}{\lim }{\int }_1^{-1}\frac{e^{-x^2}}{x^2}dx+{\int }_{-1}^0\frac{e^{-x^2}}{x^2}dx+{\int }_0^1\frac{e^{-x^2}}{x^2}dx+\underset{\theta \to \infty }{\lim }{\int }_1^8\frac{e^{-x^2}}{x^2}dx.$

Step-by-step solution

Step 1. Given information.   

We are given an integrals,

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

Step 2. Graph of function 

Let $f\left(x\right)=\frac{e^{-x^2}}{x^2}$,

The graph is as follows,

Step 3. Expressing the Integral. 

The function $f\left(x\right)=\frac{e^{-x^2}}{x^2}$is continuous everywhere except at $x=0$.

Thus, split the interval $(-\infty ,\infty )$at $x=0$.

Rewriting the improper integral by splitting the interval $(-\infty ,\infty )$at $x=0$.

${\int }_{-\infty }^{\infty }\frac{e^{-x^2}}{x^2}dx={\int }_{-\infty }^0\frac{e^{-x^2}}{x^2}dx+{\int }_0^{\infty }\frac{e^{-x^2}}{x^2}dx$

Further, split the intervals at the points $x=\pm 1$to make the calculations of the integrals more simple way.

Hence, split the interval $(-\infty ,\infty )$as the following subintervals:

$(-\infty ,-1),(-1,0),(0,1)\text{and}(1,\infty )\text{.}$

Therefore, the improper integral becomes,

$\underset{x\to -}{\lim }{\int }_1^{-1}\frac{e^{-x^2}}{x^2}dx+{\int }_{-1}^0\frac{e^{-x^2}}{x^2}dx+{\int }_0^1\frac{e^{-x^2}}{x^2}dx+\underset{\theta \to \infty }{\lim }{\int }_1^8\frac{e^{-x^2}}{x^2}dx.$