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I am trying to compute

$$ \int_0^\infty \frac{\ln x}{x^2 +4}\,dx,$$

which I find here, without complex analysis. I am consistently getting the wrong answer and am hoping someone can spot the error.

First, denote the integral by $I$, and take $x = \frac{1}{t}$. Hence, the integral becomes

$$I = \int_\infty^0 \frac{\ln(1/t)}{1/t^2 + 4} \left(-\frac{1}{t^2} dt \right) = \int_0^\infty \frac{\ln(1)}{1+4t^2} dt - \int_0^\infty \frac{\ln(t)}{1+4t^2} dt$$

Note that the leftmost integral on the right-hand side is zero. Now, letting $u = 2t$, we get

$$I = -\frac{1}{2} \int_0^\infty \frac{\ln(2u)}{1+u^2}\,du = - \frac{1}{2} \int_0^\infty \frac{\ln(2)}{1+u^2}\,du - \frac{1}{2} \int_0^\infty \frac{\ln(u)}{1+u^2}\,du = - \frac{1}{2} \int_0^\infty \frac{\ln(2)}{1+u^2} - \frac{I}{2}$$

and therefore $I = - \frac{\ln(2)}{3} \int_0^\infty \frac{1}{1+u^2}du = - \frac{\pi \ln(2)}{6}$. This, however, is not the right answer. So, where did I go wrong?

Community
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user193319
  • 7,940
  • Where does come from -I/2 ? BTW, $\displaystyle \int_0^{\infty} \dfrac{\ln x}{1+x^2}dx=0$ – FDP Mar 07 '17 at 15:13
  • @FDP Oh, shoot! I was thinking $\int_0^\infty \frac{\ln(u)}{1+u^2}$ was just the original integral and hence substituted in $I$! – user193319 Mar 07 '17 at 15:28

3 Answers3

6

It is probably easier to perform the following manipulations: $$ \int_{0}^{+\infty}\frac{\log x}{x^2+4}\,dx\stackrel{x\mapsto 2z}{=} 2\int_{0}^{+\infty}\frac{\log(2)+\log(z)}{4+4z^2}\,dz \\= \color{red}{\frac{\log(2)}{2}\int_{0}^{+\infty}\frac{dz}{1+z^2}}+\color{blue}{\frac{1}{2}\int_{0}^{+\infty}\frac{\log z}{1+z^2}\,dz}$$ leading to $\color{red}{\frac{\pi\log(2)}{4}}$, since the blue integral vanishes by the substitution $z\mapsto\frac{1}{z}$.

Jack D'Aurizio
  • 353,855
2

There is a mistake when writing, with $u=2t$, that $$ - \int_0^\infty \frac{\ln(t)}{1+4t^2} dt=-\frac{1}{2} \int_0^\infty \frac{\ln(2u)}{1+u^2}\:du $$ since $t=u/2$, we don't have $\ln (t)=\ln (2u).$

Olivier Oloa
  • 120,989
1

One problem is, when you substitute $u=2t$ then $\log t = \log(u/2)$, not $\log(2u)$ as you have stated in your equation $$I = -\frac{1}{2} \int_0^\infty \frac{\ln(2u)}{1+u^2}\,du$$

PM.
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