Evaluate: $\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx.$

Question

Prove that:$$\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx=\frac{2\ln c}{\sqrt{4c^2-b^2}}\cot^{-1}\left(\frac{b}{\sqrt{4c^2-b^2}}\right),$$ where $4c^2-b^2>0, c>0.$

We have: \begin{align} 4(x^2+bx+c^2)&=(2x+b)^2-\left(\underbrace{\sqrt{4c^2-b^2}}_{=k\text{ (say)}}\right)^2\\\\ &=(2x+b+k)(2x+b-k). \end{align} Thus, \begin{align} I&=4\int_0^{\infty}\frac{\ln x}{(2x+b+k)(2x+b-k)}\,dx\\\\ &=\frac4{2k}\int_0^{\infty}\left(\frac1{2x+b-k}-\frac1{2x+b+k}\right)\ln x\, dx\\\\ &=\frac2k\left[\underbrace{\int_0^{\infty}\frac{\ln x}{2x+b-k}\,dx}_{=I_1}-\underbrace{\int_0^{\infty}\frac{\ln x}{2x+b+k}\,dx}_{=I_2}\right] \end{align} For $I_1:$ Letting $2x+b-k=t$ yields \begin{align} I_1&=\int_{b-k}^{\infty} \frac{\ln(t-b+k)-\ln 2}t\,dt\\\\ &=\int_{b-k}^{\infty}\frac{\ln(t-b+k)}t\, dt-\ln 2\int_{b-k}^{\infty}\frac{dt}t \end{align} Now $\ln(t-b+k)=\ln(k-b)+\ln\left(\frac{t}{k-b}+1\right).$ So,$$I_1=\ln(k-b)\int_{b-k}^{\infty}\frac{dt}t+\int_{b-k}^{\infty} \frac{\ln\left(\frac t{k-b}+1\right)}t\,dt-\ln 2\int_{b-k}^{\infty}\frac{dt}t.$$

If we let: $\frac t{k-b}=-u.$ Then $$I_1=\ln\left(\frac{k-b}2\right)\int_{b-k}^{\infty} \frac{dt}t+\int_1^{\infty}\frac{\ln(1-u)}u\,du.$$

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It's getting messier and messier. Also it seems that $I_1$ and hence $I$ diverges. Please tell me, am I heading towards something useful? It has become difficult for me to handle this integral. Please show me a proper way.

Answer 1

Hint : Substitute $xy=c^2\to x=\frac{c^2}{y}$ then the integral becomes $$ I= \int_0^{\infty}\frac{\ln(c^2)-\ln y}{y^2+by+c^2}dy\\2I= \int_0^{\infty}\frac{\ln(c^2)}{y^2+by+c^2}dy$$ and since $$ y^2+by+c^2= \left(y+\frac{b}{2}\right)^2+\left(c^2-\frac{b^2}{4}\right)$$ and using elementary integral formula of $$\int_{0}^{\infty}\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)$$ we have our desired closed form.

Answer 2

Hint:

$$ x^2 + bx+ c^2 = (x+b/2)^2 + (c^2 - b^2/4) $$

with $c^2 - b^2/4 > 0$.

This invites change of variables using $\arctan$ as

$$d\arctan x = \frac{1}{x^2+1}dx $$

Answer 3

I will just show how to solve it in the case $b=0$, since the idea is the same, and the calculation becomes easier. Substituting $x=c\tan\theta$, we have:

\begin{equation} \begin{split} \int_0^\infty\frac{\ln x}{x^2+c^2}dx&=c^{-1}\int_0^{\pi/2}\ln(c\tan\theta)d\theta\\ &=c^{-1}\int_0^{\pi/2}\ln(c)d\theta+\int_0^{\pi/2}\ln{\sin\theta}d\theta-\int_0^{\pi/2}\ln{\cos\theta}d\theta\\ &=c^{-1}\int_0^{\pi/2}\ln(c)d\theta\\ &=c^{-1}\frac{\pi}2\ln(c)\\ \end{split} \end{equation}

Answer 4

$$ \int_0^{\infty}{\frac{\ln x}{x^2+bx+c^2}\text{d}x}=\frac{\ln c}{c^2}\int_0^{\infty}{\frac{1}{t^2+bct+1}\text{d}t}+\frac{1}{c^2}\int_0^{\infty}{\frac{\ln t}{t^2+bct+1}\text{d}t}=\frac{\ln c}{c^2}\mathcal{I}+\frac{1}{c^2}\mathcal{J} $$ $$ \mathcal{I}=\int_0^{\infty}{\frac{1}{\left( t+\frac{1}{2}bc \right) ^2+1-\frac{1}{4}b^2c^2}\text{d}t}=\frac{1}{\sqrt{1-\frac{1}{4}b^2c^2}}\arctan \frac{t+\frac{1}{2}bc}{\sqrt{1-\frac{1}{4}b^2c^2}}\mid_{0}^{\infty}=\frac{1}{\sqrt{1-\frac{1}{4}b^2c^2}}\left( \frac{\pi}{2}-\arctan \frac{bc}{\sqrt{4-b^2c^2}} \right) $$ $$ \mathcal{J}=\int_0^{\infty}{\frac{\ln t}{t^2+bct+1}\text{d}t}=-\int_0^{\infty}{\frac{\ln t}{t^2+bct+1}\text{d}t}=0 $$ $$ \int_0^{\infty}{\frac{\ln x}{x^2+bx+c^2}\text{d}x}=\frac{\ln c}{c^2}\mathcal{I}+\frac{1}{c^2}\mathcal{J}=\frac{2\ln c}{c^2\sqrt{4-b^2c^2}}\arctan \frac{\sqrt{4-b^2c^2}}{bc} $$