5

I want to solve the integral $\int_1^\infty\frac{a}{a^2+x^2}dx$ where $a>0$ is a constant

I first tried simplifying so \begin{align} \frac{a}{a^2+x^2} &=\frac{1}{a+\frac{x^2}{a}} \\[6px] &=\frac{1}{a}\cdot\frac{1}{1+(\frac{x}{a})^2} \end{align} which got me \begin{align} \int_1^\infty\frac{a}{a^2+x^2}dx &=\frac{1}{a}\int_1^\infty\frac{1}{1+(\frac{x}{a})^2}dx\\[6px] &=\frac{1}{a}\left[\arctan \frac{x}{a}\right]_1^\infty \end{align} which is wrong, but I can't see my mistake, can anyone help me with this?

egreg
  • 238,574
arnoldschwarzenegger
  • 53
  • 1
    Try differentiating the result to see if you get the integrand -- be sure to apply the chain rule! Alteratively: do the substitution $x = au, dx = a~ du$ (being sure to change the limits of integration as well). – John Hughes Jun 24 '17 at 17:11
  • 3
    Yeah, the mistake is that $\frac{d}{dx} (\arctan \frac{x}{a}) = \frac{1}{a}*\frac{1}{1+x^2/a^2}$ due to the chain rule. Therefore, you don't need the additional factor. – John Lou Jun 24 '17 at 17:11
  • 2
    That is to say, the antiderivative of $\frac{1}{a}*\frac{1}{1+(\frac{x}{a})^2}$ is $\arctan(x/a)$, which is the more common antiderivative at $a= 1$ – John Lou Jun 24 '17 at 17:16
  • I see it now, thank you! – arnoldschwarzenegger Jun 24 '17 at 17:21

1 Answers1

4

\begin{align} \int_1^\infty\frac{a}{a^2+x^2}dx &=\frac{1}{a}\int_1^\infty\frac{1}{1+(\frac{x}{a})^2}dx\\[6px] &=\int_1^\infty\frac{1}{1+(\frac{x}{a})^2}d\frac{x}{a}\\[6px] &=\left[\arctan \frac{x}{a}\right]_1^\infty\\[6px] &=\frac{\pi}{2}-\arctan\frac{1}{a}=\arctan a \end{align}

If $a<0$, then the final step will give $\pi+\arctan a$.

egreg
  • 238,574