Question

${\int }_0^{\infty }\frac{lnx}{x^2+a^2}dx$=

• A. $\left(\frac{lna}{a}\right)\frac{\pi }{4}$
• B. $0$
• C. $\left(\frac{lna}{a}\right)\frac{\pi }{3}$
• D. $\left(\frac{lna}{a}\right)\frac{\pi }{2}$

Answer 1

The correct option is D $\left(\frac{lna}{a}\right)\frac{\pi }{2}$

${\int }_0^{\infty }\frac{\ln x}{x^2+a^2}dx$

$\Rightarrow \frac{\left(\ln \right(-ia\left)\ln \right(|x+ia|)-\ln (ia\left)\ln \right(|x-ia|)+L{i}_2\left(\frac{cx+a}{a}\right)-l{i}_2(-ix-a/a\left)\right)+c}{2a}$

It is assumed that $a\ne 0$

$\frac{\pi \ln (x^2+a^2)+2il{i}_2(ix+a/a)-2il{i}_2(-ix-a/a)-4arc\tan 2(x,a)\ln \left(x/|a|\right)-l{i}_{2}arc\tan 2(0,a/|a|)arc\tan \left(2\right(x,a\left)\right)}{a}$

$+\frac{\ln \left(x\right)arc\tan \left(x/a\right)}{a}+c$

$\Rightarrow \frac{\pi \ln \left(a\right)}{2a}\Rightarrow \left(\frac{\ln a}{a}\right)\frac{\pi }{2}$