Integrals Involving Quadratics — Question 10

Question 10 – Solution

The antiderivative below has finite limits at both $-\infty$ and $\infty$, so both one-sided improper integrals converge separately. Their sum may therefore be computed using symmetric limits. Step 1: Write the improper integral as a limit. $$I=\lim_{R\to\infty}\int_{-R}^{R}\frac{dx}{x^2+a^2}.$$ Step 2: Find an antiderivative. Since $a>0$, $$\int\frac{dx}{x^2+a^2}=\frac1a\arctan\left(\frac{x}{a}\right)+C.$$ Therefore, $$\begin{align*} I&=\lim_{R\to\infty}\frac1a \left[\arctan\left(\frac{x}{a}\right)\right]_{-R}^{R}\\ &=\frac1a\lim_{R\to\infty} \left[\arctan\left(\frac{R}{a}\right) -\arctan\left(\frac{-R}{a}\right)\right]. \end{align*}$$ Step 3: Evaluate the limits. $$\begin{align*} I&=\frac1a\left(\frac\pi2-\left(-\frac\pi2\right)\right)\\ &=\frac\pi a. \end{align*}$$ $$\boxed{I=\frac\pi a}$$