# Evaluate the improper iterated integral \int\limits_0^4 \int\limits_0^{\infty}...

## Question:

Evaluate the improper iterated integral

{eq}\int\limits_0^4 \int\limits_0^{\infty} \frac{x^2}{1+y^2} \text{d}x\text{d} y {/eq}

## Double Integrals:

The inner integral is solved in the order as given in the problem, such that the final result of the double integral is obtained. But if the inner integral is diverging, the outer integral is diverging and overall integral too is diverging.

The double integral given here is:

{eq}\int_0^4 \int_0^{\infty} \frac{x^2}{1+y^2} dx dy\\ {/eq}

In this integral first the inner order wrt dx is to be evaluated as follows:

{eq}\Rightarrow \int_0^{\infty} \frac{x^2}{1+y^2}dx\\ =\left [ \frac{x^3}{3\left(1+y^2\right)} \right ]_0^{\infty}~~~~~~~~~~~~~~~\left [ \because \int x^adx=\frac{x^{a+1}}{a+1}+c \right ]\\ =\lim _{x\to \infty \:}\left(\frac{x^3}{3\left(1+y^2\right)}\right)-\lim _{x\to \:0+}\left(\frac{x^3}{3\left(1+y^2\right)}\right)\\ =\infty \:\\ {/eq}

So here we have the diverging inner part and hence the outer integral too will be diverging:

{eq}\Rightarrow \int_0^4 \int_0^{\infty} \frac{x^2}{1+y^2} dx dy= \infty {/eq} 