Compute $$ \int_{0}^{ \infty }\frac{\ln xdx}{x^{2}+ax+b}$$ I tried to compute the indefinite integral by factorisation and partial fraction decomposition but it became nasty pretty soon. There must be another way to directly evaluate it without actually computing the indefinite integral which I don't know!
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Have you tried the substitution $ln x =t$? – john doe Oct 15 '17 at 09:06
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In order for your integral to converge, we must have $b>0$.
Hence $$\begin{aligned}\int_0^\infty {\frac{{\ln x}}{{{x^2} + ax + b}}dx} &= \sqrt b \int_0^\infty {\frac{{\ln (\sqrt b x)}}{{b{x^2} + a\sqrt b x + b}}dx} \\ &= \sqrt b \ln \sqrt b \int_0^\infty {\frac{1}{{b{x^2} + a\sqrt b x + b}}dx} + \sqrt b \underbrace{\int_0^\infty {\frac{{\ln x}}{{b{x^2} + a\sqrt b x + b}}dx}}_{=0} \end{aligned}$$
The last integral is zero by making the subsitution $x\mapsto 1/x$. I think you will have no difficulty in evaluating the remaining integral. Although there are a lot of cases to discuss (depending on $a,b$, the integral might diverge, involve logarithm, or involve arc-tangent).
pisco
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1This one reminds me of Jack's answer: https://math.stackexchange.com/a/2463284/72031 +1 – Paramanand Singh Oct 15 '17 at 13:25
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Let $$I = \int^{\infty}_{0}\frac{\ln x}{x^2+ax+b}dx$$
Put $\displaystyle x = \frac{b}{t}$ and $\displaystyle dx = -\frac{b}{t^2}dt$
So $$I = \int^{\infty}_{0}\frac{\ln (b)-\ln(t)}{t^2+at+b}dt = \ln (b)\int^{\infty}_{0}\frac{1}{t^2+at+b}dt-I$$
So $$I = \frac{\ln (b)}{2}\int^{\infty}_{0}\frac{1}{\left(t+\frac{a}{2}\right)^2+\left(b-\frac{a^2}{4}\right)}dt = \frac{\ln (b)}{2}\cdot \frac{4}{4b-a^2}\cdot \tan^{-1}\left(\frac{4t+2a}{4b-a^2}\right)\Bigg|^{\infty}_{0}$$
for $\displaystyle b-\frac{a^2}{4}>0$
DXT
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