Evaluate \[ \int_0^\infty\frac{\log x}{(x^2+1)(x^2+4)}\,dx. \]
Solution
Let \[ I=\int_0^\infty\frac{\log x}{(x^2+1)(x^2+4)}\,dx \] and, for \(a>0\), let \[ J(a)=\int_0^\infty\frac{\log x}{x^2+a^2}\,dx. \]
1. One has\[ I=\frac13\bigl(J(1)-J(2)\bigr). \]
The partial-fraction identity \[ \frac1{(x^2+1)(x^2+4)} = \frac13\left(\frac1{x^2+1}-\frac1{x^2+4}\right) \] gives the claim after integration.
2. The integral\[ K=\int_0^\infty\frac{\log t}{1+t^2}\,dt \]converges and satisfies \(K=0\).
Near \(0\), the absolute value is bounded by \(\abs{\log t}\), whose integral over \((0,1]\) is finite. For \(t\geq1\), \[ \frac{\log t}{1+t^2}\leq\frac{\log t}{t^2}, \] whose integral over \([1,\infty)\) is finite. Thus \(K\) is absolutely convergent.
Split the integral at \(1\). In the first part, substitute \(t=1/u\): \[ \begin{aligned} \int_0^1\frac{\log t}{1+t^2}\,dt &= -\int_1^\infty\frac{\log u}{1+u^2}\,du. \end{aligned} \] Therefore the integrals over \((0,1)\) and \((1,\infty)\) cancel, so \(K=0\).
3. For every \(a>0\),\[ J(a)=\frac{\pi\log a}{2a}. \]
2. Hence\[ J(a)=\frac{\pi\log a}{2a}. \]
4. The required value is\[ \boxed{ \int_0^\infty\frac{\log x}{(x^2+1)(x^2+4)}\,dx =-\frac{\pi\log2}{12} }. \]
By step 3, \[ J(1)=0, \qquad J(2)=\frac{\pi\log2}{4}. \] Substituting these values into step 1 gives \[ I = \frac13\left(0-\frac{\pi\log2}{4}\right) = -\frac{\pi\log2}{12}. \]
5. Q.E.D.
Step 4 is the required evaluation.