Showing that the integrals $\int_{\epsilon}^1\frac{\ln(x)}{x^4+1}dx$ and $\int_{1}^{\infty}\frac{\ln(x)}{x^4+1}dx$ converge as $\epsilon\downarrow 0$ My complex analysis book says quite bluntly that: "It is not difficult to show that the integrals $\int_{\epsilon}^1\frac{\ln(x)}{x^4+1}dx$  and $\int_{1}^{\infty}\frac{\ln(x)}{x^4+1}dx$ converge as $\epsilon\downarrow 0$". Mathematica's reaction to the first integral is Undefined
and I unfortunately do not know any good trick to bound the $\frac{\ln(x)}{x^4 + 1}$. Therefore I am asking 1.) how could one show the claimed convergence, 2.) is there a general approach for these sort of limit integrals involving hard-to-integrate functions, such as logarithms?
 A: The simple integral comparison test: If $f(x)\leq g(x)$ on the interval $(a,b)$, then $\int_a^b f(x)dx\leq\int_a^b g(x)dx$.
Since, $\ln x<0$ and $\frac{1}{2}\leq\frac{1}{x^4+1}\leq 1$ on $(0,1)$, we have
$$\ln x\leq \frac{\ln x}{x^4+1}\leq \frac{1}{2}\ln x$$
and
$$\int_0^1 \ln x dx=-1\leq\int_0^1 \frac{\ln x}{x^4+1}dx\leq -\frac{1}{2}=\int_0^1 \frac{1}{2}\ln x dx.$$
So, $\int_0^1 \frac{\ln x}{x^4+1}dx=\lim_{\epsilon\rightarrow 0^+}\int_{\epsilon}^1 \frac{\ln x}{x^4+1}dx$ is convergent. WA says it is $\approx -0.9685$.
The convergence test for the other integral is easier. Since, $0<\int_1^\infty \frac{\ln x}{x^4+1}dx\leq \int_1^\infty \frac{2x}{x^4+1}dx=\frac{\pi}{4}$, it is convergent. Wa says, $\int_1^\infty \frac{\ln x}{x^4+1}dx\approx 0.0961$.
A: Usually you can't compute the integral , and it's generally not needed in order to show convergence. You can compare it to usual fonctions like $x^\alpha$ at the problemetic points ($0$ and +$\infty$) which is the most common method.
$\frac{\ln(x)}{x^4+1} \sim_{0} \ln(x) = o(\frac{1}{\sqrt{x}})$ and $\frac{1}{\sqrt{x}}$ is integrable in the neighborhood of $0$.
Fort the second one, $\frac{\ln(x)}{x^4+1} \sim_{+\infty} \frac{\ln(x)}{x^4}= o(\frac{1}{x^3}) $ which is integrable in the neighborhood of infinity.
A: For the integral in $[\epsilon,1]$, notice that $x^4+1>1$, it is $\left|\frac{\log x}{x^4+1}\right|<|\log x|$.
Similarly, for the integral in $[1,\infty)$ since it is $\log x>0$ for $x>1$ and $\log x<x$ for $x>0$, you can estimate $\frac{\log x}{x^4+1}<\frac{x}{x^4}=\frac{1}{x^3}$.
The estimation $\log(1+t)<t$ for any $t>-1$ is usually useful in these contexts.
