I have been trying to evaluate or at least get a tight upper bound on the definite integral
$$ \int_{0}^c \sqrt{\frac{1}{x} \log\left(1+\frac{1}{x}\right)} dx$$
for some $c>1$. The integral converges but the standard estimate $\log(1+\frac{1}{x}) \leq \frac{1}{x}$ would not work, as $\int_{0}^c \frac{dx}{x} $ diverges. Any help would be greatly appreciated.
Best Answer
Too long for a comment $$I(c)=\int_0^c \sqrt{\frac{1}{x} \log\left(1+\frac{1}{x}\right)} dx\overset {x=\frac1t}{=}\int_{1/c}^\infty t^{-3/2}\sqrt{\ln(1+t)}\,dt$$ $$\overset{IBP}{=}\ln t\,\sqrt{\frac{\ln(1+t)}{ t}}\bigg|_{1/c}^\infty-\int_{1/c}^\infty\ln t\left(\sqrt{\frac{\ln(1+t)}{ t}}\right)^{'}dt$$ $$=\ln c\sqrt{c\ln\big(1+\frac1c\big)}-\int_0^\infty\ln t\left(\sqrt{\frac{\ln(1+t)}{t}}\right)^{'}dt+\int_0^{1/c}\ln t\left(\sqrt{\frac{\ln(1+t)}{ t}}\right)^{'}dt$$ The second integral converges; I'm not sure that the closed form exists.
Numerically $$-\int_0^\infty\ln t\left(\sqrt{\frac{\ln(1+t)}{ t}}\right)^{'}dt=\frac12\int_0^\infty\frac{\ln t}{\sqrt{t\ln(1+t)}}\left(\frac{\ln(1+t)}t-\frac1{1+t}\right)dt=2.6256...$$ The third integral can be evaluated via the decomposition of the integrand $(\frac1c<1)$: $$\int_0^{1/c}\ln t\left(\sqrt{\frac{\ln(1+t)}{t}}\right)^{'}dt=-\frac14\int_0^{1/c}\frac{\ln t}{\sqrt{1-\frac t2+\frac{t^2}3+...}}\left(1-\frac {4t}3+...\right)dt$$ $$=\frac{\ln c+1}{4c}-\frac{13}{192}\frac{2\ln c+1}{c^2}+O\Big(\frac{\ln c}{c^3}\Big)$$ Finally, $$I(c)=\ln c\sqrt{c\ln\Big(1+\frac1c\Big)}+2.6256...+\frac{\ln c+1}{4c}-\frac{13}{192}\frac{2\ln c+1}{c^2}+O\Big(\frac{\ln c}{c^3}\Big)$$
Numeric check:
$\,c=4\quad I(4)=4.07058...\,;\quad\text{approximation}=4.06850...$
$\,c=10\quad I(10)=4.95253...\,;\quad\text{approximation}=4.95231...$