I want to evaluate the integration $$I(\epsilon)=\int_0^\infty \frac{1}{\sqrt {x(x+\epsilon)(1+x^5)}}\,dx \quad(\epsilon\to0)$$
Although $I(0)$ is divergent, one may still derive an approximate expression for $I(\epsilon)$ at small $\epsilon$ or a numerical expression with $\epsilon$ being a parameter.
Using Mathematica's LogLinearPlot to draw a logarithmatic graph,I found that $I(\epsilon)$ is linear to $\epsilon$'s logarithm. And by linear fitting I made a guess: When $\epsilon$ goes to $0$, there will be the relation $$I(\epsilon)= -\ln\epsilon+\textrm{const.}$$
But how can I figure out the constant above? By assymptotic expansion or something?
Or the observation and assumption I made are wrong as a whole? Is there any other way to evaluate the approximate or numerical expression of $I(\epsilon)?$
Integrate by parts, writing $\displaystyle\frac{1}{\sqrt{x(x+\epsilon)}}=\frac{d}{dx}\ln\color{gray}{\frac{1}{2}}\left(x+\frac{\epsilon}{2}+\sqrt{x(x+\epsilon)}\right)$. As a result, you get $$\lim_{\epsilon\to+0}\left(\ln\epsilon+\int_0^\infty\frac{dx}{\sqrt{x(x+\epsilon)(1+x^a)}}\right)=2\ln 2+\frac{a}{2}\int_0^\infty\frac{x^{a-1}\ln x}{(1+x^a)^{3/2}}\,dx=2\ln 2+\frac{A}{2a}$$ (for real $a>0$) where $A=\displaystyle\int_0^\infty\frac{\ln t\ dt}{(1+t)^{3/2}}=4\ln 2$ (integrated elementarily).
Thus the limit is equal to $2(1+1/a)\ln 2$. Put $a=5$ to get your "$\mathrm{const.}$".