- Show that if \(a > 0\) , then
\[ \int_ {0} ^ {\infty} \frac {\log x}{x ^ {2} + a ^ {2}} d x = \frac {\pi}{2 a} \log a. \]
[Hint: Use the contour in Figure 10.]
Solution
Put \(x=at\). Then \[ I(a):=\int_0^\infty \frac{\log x}{x^2+a^2}\,dx =\frac1a\int_0^\infty \frac{\log a+\log t}{1+t^2}\,dt. \] Now \[ \int_0^\infty \frac{dt}{1+t^2}=\frac\pi2. \] Also, with \(t=1/u\), \[ \int_0^\infty \frac{\log t}{1+t^2}\,dt =\int_\infty^0 \frac{-\log u}{1+u^{-2}}\frac{-du}{u^2} =-\int_0^\infty \frac{\log u}{1+u^2}\,du, \] so this integral is \(0\). Therefore \[ I(a)=\frac1a\left(\log a\cdot\frac\pi2\right)=\frac{\pi}{2a}\log a. \]