Asymptotics of an integral involving the exponential integral

Question

Consider the integral: $$ I(a)=\int_a^\infty e^x E_1(x)\dfrac{dx}{x}, $$ where $a>0$ and $E_1(x)$ is the exponential integral function.

I would like to better understand the behavior of $I(a)$ for $a\to 0+$. For example, from the double inequality $$ \dfrac{1}{x+1}<e^{x}E_1(x)<\dfrac{1}{x},\ \ x>0, $$ we readily get $$ \log(1+1/a)<I(a)<\dfrac{1}{a}. $$

The (more fine) double inequality $$ \dfrac{1}{2}\log(1+2/x)<e^{x}E_1(x)<\log(1+1/x),\ \ x>0, $$ yields $$ 1/12 (\pi^2 + 3 \log^2(2) + 3 \log(a/4)\log(a) + 6 Li_2(-a/2))<I(a)<-Li_2(-1/a), $$ where $Li_2(x)$ is the poly-log function.

Can these bounds be refined? For example, are there tight bounds for the poly-log function?

Answer 1

$$I(a)=\int_a^\infty e^x E_1(x)\dfrac{dx}{x}=\int_a^\infty \frac{e^x}xdx\left(\int_x^\infty\frac{e^{-t}}tdt\right)\overset{t=xs}{=}\int_a^\infty \frac{e^x}xdx\left(\int_1^\infty\frac{e^{-xs}}sds\right)$$ $$\overset{x=at}{=}\int_1^\infty \frac{e^{at}}tdt\left(\int_1^\infty\frac{e^{-ats}}sds\right)=\int_1^\infty\frac{dt}t\int_0^\infty\frac{e^{-ats}}{1+s}ds$$ Integrating by part $$=\int_1^\infty\frac{dt}t\left(\ln(1+s)e^{-ats}\,\bigg|_0^\infty+at\int_0^\infty\ln(1+s)e^{-ats}ds\right)=\int_0^\infty\frac{\ln(1+s)}se^{-as}ds\tag{1}$$ Making the substitution $as=t$ and integrating by part $$=\ln s\ln\frac{s+a}ae^{-s}\,\bigg|_0^\infty-\int_0^\infty\frac{\ln s}{a+s}e^{-s}ds+\int_0^\infty\ln s\ln\frac{a+s}ae^{-s}ds$$ $$=\int_0^\infty\ln s\ln(a+s)e^{-s}ds+\gamma\ln a-\int_0^\infty\frac{\ln s}{a+s}e^{-s}ds$$ The first integral converges at $a=0$; therefore $$I(a)=\gamma\ln a+\int_0^\infty\ln^2s\,e^{-s}ds-\int_0^\infty\frac{\ln s}{a+s}e^{-s}ds+o(1)\tag{2}$$ Using $\displaystyle \frac1{s+a}=\int_0^\infty e^{-t(s+a)}dt$, the last integral in (2) can be evaluated. Changing the order of integration, $$-\int_0^\infty\frac{\ln s}{a+s}e^{-s}ds=-\int_0^\infty e^{-ta}dt\int_0^\infty\ln s\,e^{-s(t+1)}ds$$ $$=-\int_0^\infty\frac{dt}{1+t}e^{-at}\left(-\gamma-\ln(1+t)\right)=e^a\int_1^\infty\frac{\ln x}xe^{-ax}dx+\gamma e^a\int_1^\infty\frac{e^{-ax}}xdx$$ Integrating by part and keeping only non-vanishing at $a\to 0$ terms, we get $$-\int_0^\infty\frac{\ln s}{a+s}e^{-s}ds=\frac{e^a}2\int_a^\infty(\ln x-\ln a)^2e^{-x}dx-\gamma\ln a+\gamma e^a\int_a^\infty\ln xe^{-x}dx$$ $$=\frac{\ln^2a}2+\frac12\int_0^\infty\ln^2xe^{-x}dx-\gamma^2+o(1)=\frac{\ln^2a}2+\frac{\pi^2}{12}-\frac{\gamma^2}2+o(1)\tag{3}$$ Putting (3) into (2) $$\boxed{\,\,I(a)=\frac{\ln^2a}2+\gamma\ln a+\frac{\gamma^2}2+\frac{\pi^2}4+o(1)\,\,}$$

Answer 2

$$I(a)=\int_a^\infty\frac{ e^x} x E_1(x)\,dx$$ As usual, the result is given in terms of hypergeometric functions and some of their derivatives $$I(a)= \text{Ei}(a) \,(\log (a)+\gamma )-\frac 12 \log (a) \, (\log (a)+2 \gamma )+\frac{(\pi ^2-2 \gamma ^2)}4-$$ $$2 a \, _3F_3(1,1,1;2,2,2;a)-\frac{1}{2}\, _1F_1^{(2,0,0)}(0;1;a)$$

Expanded as series around $a=0^+$

$$\color{blue}{I(a)=\left(\gamma +a+\frac{a^2}{4}+\frac{a^3}{18} \right)\,\log(a)+\frac{1}{2}\log ^2(a)+}$$ $$\color{blue}{\frac{2 \gamma ^2+\pi ^2}{4}+(\gamma -2) a+\frac{\gamma -2}{4} a^2+\frac{6 \gamma -13}{108} a^3}$$ which is a very good approximation up to $a=1$.

Just to give an idea $$\int_0^{\frac 12} \Big(I(a)-\text{approximation}\Big)^2\,da=1.81 \times 10^{-7}$$ $$\int_0^1 \Big(I(a)-\text{approximation}\Big)^2\,da=5.84 \times 10^{-5}$$