Let \(a>0\). Evaluate \[ \int_0^\infty\frac{\log x}{a^2+x^2}\,dx. \]
Solution
1. The improper integral converges absolutely.
On \((0,1]\), \[ \frac{\abs{\log x}}{a^2+x^2} \leq \frac{\abs{\log x}}{a^2}, \] and \[ \int_0^1\abs{\log x}\,dx=1. \] On \([1,\infty)\), \[ \frac{\log x}{a^2+x^2} \leq \frac{\log x}{x^2}, \] and integration by parts gives \[ \int_1^\infty\frac{\log x}{x^2}\,dx=1. \] Hence the given integral is absolutely convergent.
2. With\[ J=\int_0^\infty\frac{\log t}{1+t^2}\,dt, \]one has\[ \int_0^\infty\frac{\log x}{a^2+x^2}\,dx = \frac{\pi\log a}{2a} + \frac{J}{a}. \]
Set \(x=at\). Since \(a>0\), \[ dx=a\,dt \] and \[ a^2+x^2=a^2(1+t^2). \] Thus \[ \begin{aligned} \int_0^\infty\frac{\log x}{a^2+x^2}\,dx &= \frac1a \int_0^\infty \frac{\log a+\log t}{1+t^2}\,dt\\ &= \frac{\log a}{a} \int_0^\infty\frac{dt}{1+t^2} + \frac{J}{a}. \end{aligned} \] Since \[ \int_0^\infty\frac{dt}{1+t^2} = \left[\arctan t\right]_0^\infty = \frac\pi2, \] the claimed formula follows.
3. One has\[ J=0. \]
Split \[ J = \int_0^1\frac{\log t}{1+t^2}\,dt + \int_1^\infty\frac{\log t}{1+t^2}\,dt. \] In the second integral, set \(t=1/u\). Then \[ dt=-\frac{du}{u^2}, \] and therefore \[ \int_1^\infty\frac{\log t}{1+t^2}\,dt = -\int_0^1\frac{\log u}{1+u^2}\,du. \] This cancels the first integral, so \(J=0\).
4. Therefore\[ \boxed{ \int_0^\infty\frac{\log x}{a^2+x^2}\,dx = \frac{\pi\log a}{2a} }. \]
Substitute step 3 into the formula of step 2.
5. Q.E.D.
Step 4 gives the requested value.