Bound an integral with a parameter

Question

I am reading a paper and at some point they reach the following integral $$f(t) = \int_0^\infty\frac{8x^3}{(1 + x^2)^3} \frac{1 - e^{-tx^2/4}}{tx^2}dx=\int_0^\infty\frac{8x(1 - e^{-tx^2/4})}{t(1 + x^2)^3}dx.$$ From that they claim that \begin{align} |f(t) - 1| &\le C t(1 + |\log t|) \quad \text{for } t < 1. \end{align} I've tried many things but I couldn't find the same bound for $f(t)$. Note that $$\int_0^\infty\frac{8x^3}{(1 + x^2)^3} dx = 2.$$ Therefore $$f(t) - 1 = \int_0^\infty \frac{8x}{t(1 + x^2)^3}\left\{1 - \frac{tx^2}{2} - e^{-tx^2/4}\right\}dx$$ If we choose $u = \frac{tx^2}{4}$, the integral becomes \begin{align} f(t) - 1 &=\int_0^\infty \frac{8}{t(1 + 4u/t)^3}\left\{1 - 2u - e^{-u}\right\}\frac{2}{t}du\\ &= t\underbrace{\int_0^\infty \frac{16}{(t + 4u)^3}\left\{1 - 2u - e^{-u}\right\}du}_{\equiv A(t)} \end{align} By plotting $A(t)$ on Geogebra for $t \to 0$, I see that this should go to infinity, but I don't know how to show that it is bounded by $(1 + |\log t|)$. I guess I must use the fact that $t < 1$ somewhere.. Could anyone help me with this?

Answer 1

If you are familiar with the exponential integral function, there is an antiderivative.

Assuming $t>0$ (otherwise the integral does not converge) $$f(t)=\frac{1}{8} \left(4+t \,e^{t/4} \, \text{Ei}\left(-\frac{t}{4}\right)\right)$$ Expanded as a series around $t=0$ $$f(t)=\frac{1}{2}+\frac{1}{8} t\, (\log (t)+\gamma -\log (4))+$$ $$\frac{1}{32} t^2 \,(\log (t)+\gamma -1-\log (4))+O\left(t^3\right)$$

Now, you can work some inequalities for $0 \leq t \leq 1$.