Calculate $\int_0^\infty \frac{\ln x}{\sqrt{x}(x+1)} \, \mathrm{d}x$

Question

Calculate $$\int_0^\infty \frac{\ln x}{\sqrt{x}(x+1)} \, \mathrm{d}x$$

My try:

We will use pacman style path. in the pacman there is only a single singularity the outer path of the pacman is: $$ \lim_{R\rightarrow\infty} \oint_{\lim C} \frac{\ln Re^{i\theta}} {\sqrt{R}e^{\theta i}(Re^{\theta i}+1)} \, \mathrm{d} \theta=0.$$ now we have the "inside the pacman" which contains only a removable singularity. therefore: \begin{align} & \oint\limits_\text{pacman} \frac{\ln x}{\sqrt{x}(x+1)} \, dx \\[8pt] = {} & \lim_{R\rightarrow\infty} \oint_C \frac{\ln Re^{i\theta}}{\sqrt{R} e^{\theta i}(Re^{\theta i}+1)} \, d\theta + \oint_\varepsilon \frac{\ln Re^{i\theta}}{\sqrt{R}e^{\theta i}(Re^{\theta i}+1)} \, d\theta + 2\oint_\varepsilon^\infty \frac{i\ln e^{\theta}}{Re^{\frac{\theta i}{2}}(e^{\theta i}R+1)} \, dR\\[8pt] 0 = {} & 0+0+2\oint_\varepsilon^\infty \frac{i\ln e^\theta}{Re^{\frac{\theta i}{2}}(e^{\theta i}R+1)} \, dR\\[8pt] & \lim_{\varepsilon\rightarrow0} 2 \oint_\varepsilon^\infty \frac{x}{\sqrt{(x)}(x+1)} \, dR = 2\oint_0^\infty \frac{x}{\sqrt{(x)}(x+1)} \, dx=0\\[8pt] & \oint_0^\infty \frac{x}{\sqrt{(x)}(x+1)} \, dx=0 \end{align}

Answer 1

An alternative approach, to verify you got the right answer: with $u=\ln x$ your integral becomes$$\int_{\Bbb R}\frac{ue^{-u/2}}{1+e^{-u}}\mathrm{d}u,$$which has odd integrand, so is $0$ provided$$\int_0^\infty\frac{ue^{-u/2}}{1+e^{-u}}\mathrm{d}u$$is finite. Indeed, the latter integral has upper bound $\displaystyle \int_0^\infty ue^{-u/2}\mathrm{d}u=4$.

Answer 2

Your computation is wrong as you didn't used the correct branch of the complex logarithm for a pacman-like contour, that is, if we define the contour $\Gamma (r,R,\epsilon )$ by

then we need to use the branch of the logarithm such that $\arg(z)\in[0,2\pi)$ instead of the usual one where $\arg(z)\in(-\pi,\pi]$, then for $r,R,\epsilon >0$ we find that

$$ \begin{align*} \int_{\Gamma (r,R,\epsilon )}f(z)\mathop{}\!d z&=\lim_{\epsilon \to 0^+}\int_{\Gamma (r,R,\epsilon )}f(z)\mathop{}\!d z\\ &=\int_{0}^{2\pi} f(Re^{i\theta })d(Re^{i\theta })+\int_{0}^{2\pi} f(re^{i\theta })d(re^{i\theta })+\int_{r}^R f(x)\mathop{}\!d x-\int_{r}^{R}\lim_{\epsilon \to 0^+}f(x-i\epsilon )d(x- i \epsilon ) \end{align*} $$

for $f(z):=\frac{\ln z}{z^{1/2}(1+z)}$, then from the chosen branch of the logarithm we have that $\lim_{\epsilon \to 0^+}f(x-i\epsilon )=\frac{\ln x+2\pi i}{-\sqrt{x}(1+x)}$, and for $R>1$ we have that

$$ \left| \int_{0}^{2\pi} f(Re^{i\theta })d(Re^{i\theta }) \right|\leqslant 2\pi \sqrt R\frac{\ln R+2\pi}{R-1}\to 0\quad \text{ as }R\to \infty $$

Similarly for $r\in(0,1)$ we have that

$$ \left| \int_{0}^{2\pi} f(re^{i\theta })d(re^{i\theta }) \right|\leqslant 2\pi \sqrt r(\ln r+2\pi)\to 0\quad \text{ as }r\to 0^+ $$

Also from the residue theory we have that

$$ \int_{\Gamma (r,R,\epsilon )}f(z)\mathop{}\!d z=2\pi i\operatorname{Res}(f,-1)=2\pi^2 i\quad \text{ when }0<r<1<R $$

Putting all together we have that

$$ 2\pi^2 i=\int_{0}^{\infty }\frac{2\ln x+2\pi i}{\sqrt{x}(1+x)}\mathop{}\!d x\\ \therefore\quad \int_{0}^{\infty }\frac{\ln x}{\sqrt{x}(1+x)}\mathop{}\!d x=0\quad \text{ and }\quad \int_{0}^{\infty }\frac1{\sqrt{x}(1+x)}\mathop{}\!d x=\pi $$

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