A tough integral As a physicist of living matter, I've come across this integral in my work : $\int_1^x\frac{\tanh(u)}{u^4} du$. Does anyone have an idea how to compute it ? Mathematica returns nothing of value, and I need more than asymptotics...
Thanks a lot !
 A: I hope this estimation is good enough for you.
Remarking that $\tanh(u)=\sinh(u)/\cosh(u)$, and $\cosh'(u)=\sinh(u)$, one can integrate by parts obtaining
\begin{align}\int_1^x u^{-4}\tanh(u)du&=\left[u^{-4}\ln\cosh(u)\right]_{1}^{x}+\frac{1}{5}\int_{1}^x u^{-5}\ln\cosh(u)du=\\
&=\frac{\ln\cosh(x)}{x^4}-\ln\cosh(1)+\frac{1}{5}\int_{1}^x u^{-5}\ln\cosh(u)du.\end{align}
$1\le\cosh(u)=\frac{e^u+e^{-u}}{2}\le e^u$ for $u\ge1$, that is $0\le\ln\cosh(u)\le u$ for $u\ge1$.
It follows that
\begin{align}
\frac{1}{20}(1-x^{-4})\le\int_1^x u^{-4}\tanh(u)du-\frac{\ln\cosh(x)}{x^4}+\ln\cosh(1)\le\frac{1}{15}(1-x^{-3}).
\end{align}
Again, if you don't like having $\ln\cosh(x)$, you can use the same estimation as above and obtain
\begin{align}
-\ln\cosh(1)+\frac{1}{20}(1-x^{-4})\le\int_1^x u^{-4}\tanh(u)du\le x^{-3}-\ln\cosh(1)+\frac{1}{15}(1-x^{-3}).
\end{align}
EDIT: I remarked my estimation have been really sloppy, since I could have use this estimation instead
\begin{align}
\frac{e^{u}}{2}\le\cosh(u)=\frac{e^u+e^{-u}}{2}\le e^u \Rightarrow u-\ln2\le\ln\cosh(u)\le u,\ \text{for}\ u\ge1.
\end{align}
It follows
\begin{align}
F(x)-\frac{\ln(2)}{20}(1-x^{-4})\le\int_1^x u^{-4}\tanh(u)du\le F(x).
\end{align}
where $F(x)=\frac{\ln\cosh(x)}{x^4}-\ln\cosh(1)+\frac{1}{15}(1-x^{-3}).$
