Consider the following integral $$ \int_0^{\infty}\frac{e^{x}1}{x^{2+\frac{A}{\log b5/6}}}\frac{1}{\log(b/x)i\pi/2}\,dx $$ where $A>0$ and $b>0$. I am interested in the small $b$ asymptotics of this integral. Any ideas on how to proceed?
3 Answers
The integral in question is
\begin{equation*}
I:=\int_0^\infty f(x)\,dx, \tag{1}
\end{equation*}
where
\begin{equation*}
f(x):=\frac{e^{x}1}{x^{2\varepsilon}}\frac{1}{\ln(b/x)i\pi/2},\quad
\varepsilon:=\frac{A}{\ln b5/6}\downarrow0.
\end{equation*}
Let
\begin{equation*}
C:=C_\varepsilon:=A\tfrac56\,\varepsilon,
\end{equation*}
so that
\begin{equation*}
b=e^{C/\varepsilon}.
\end{equation*}
We have
\begin{equation*}
\int_{b^{1/2}}^\infty f(x)\,dx
\sim\int_{b^{1/2}}^\infty \frac{e^{x}1}{x^{2\varepsilon}}\frac{dx}{\ln(b/x)}
=\frac{J_1+J_2}{b^{1\varepsilon}},
\end{equation*}
where
\begin{equation*}
J_1:=\int_{\varepsilon^{1/2}/b}^\infty \frac{1e^{by}}{y^{2\varepsilon}}\frac{dy}{\ln y}
\le\int_{\varepsilon^{1/2}/b}^\infty \frac{dy}{y^{2\varepsilon}}\frac{1}{\ln(\varepsilon^{1/2}/b)}
\ll\varepsilon^{1/2}b^{1\varepsilon}=o(b^{1\varepsilon})
\end{equation*}
and, with $u=\varepsilon\ln y$,
\begin{equation*}
J_2:=\int_{b^{1/2}}^{\varepsilon^{1/2}/b} \frac{1e^{by}}{y^{2\varepsilon}}\frac{dy}{\ln y}
\sim b \int_{b^{1/2}}^{\varepsilon^{1/2}/b} \frac{dy}{y^{1\varepsilon}}\frac{1}{\ln y}
= b \int_{C/2}^{C+\varepsilon\ln(\varepsilon^{1/2})}\frac{du}u\,e^u
\sim b(\text{Ei}(A)\text{Ei}(A/2)).
\end{equation*}
It follows that
\begin{equation*}
\int_{b^{1/2}}^\infty f(x)\,dx
\to e^{A}[\text{Ei}(A)\text{Ei}(A/2)]. \tag{2}
\end{equation*}
Also, for
\begin{equation*}
g(x):=\frac1{x^{1\varepsilon}}\frac{1}{\ln(b/x)i\pi/2},
\end{equation*}
we have
\begin{equation*}
\int_0^{b^{1/2}} f(x)g(x)\,dx
\ll\int_0^{b^{1/2}} \frac{x^2}{x^{2\varepsilon}}\frac{dx}{\pi/2}\to0. \tag{3}
\end{equation*}
Then, using the substitution $t=\varepsilon\ln(b/x)$, we have
\begin{equation*}
\int_0^{b^{3/2}} g(x)\,dx\sim
\int_0^{b^{3/2}}\frac{dx}{x^{1\varepsilon}}\frac{1}{\ln(b/x)}
=e^{C}\int_{C/2}^\infty \frac{dt}t\,e^{t}=e^{C}\text{E}_1(C/2)\to
e^{A}\text{E}_1(A/2). \tag{4}
\end{equation*}
Further,
\begin{equation*}
\int_{b^{3/2}}^{b^{1/2}} g(x)\,dx=
\int_{b^{3/2}}^b g(x)\,dx+\int_b^{b^{1/2}} g(x)\,dx
=\int_{b^{1/2}}^1 g(bu)\,b\,du+\int_{b^{1/2}}^1 g(b/u)\,\frac{b\,du}{u^2}
=b\times\int_{b^{1/2}}^1 h(u)\,du,
\end{equation*}
where
\begin{equation*}
h(u):=g(bu)+g(b/u)/{u^2}
=2i e^{C(1/\varepsilon1)}e^t\,\frac{2\pi\cosh(\varepsilon t)+4i t \sinh(\varepsilon t)}{\pi^2+4t^2}
\end{equation*}
and $t:=\ln u$.
(The latter identity is the key: note that $\Re g(x)$, as well as $\Re f(x)$, changes sign at $x=b$.)
So,
\begin{equation*}
\int_{b^{3/2}}^{b^{1/2}} g(x)\,dx=
2i b e^{C(1/\varepsilon1)}\int_0^{C/(2\varepsilon)}dt\,\frac{2\pi\cosh(\varepsilon t)+4i t \sinh(\varepsilon t)}{\pi^2+4t^2}. \tag{5}
\end{equation*}
Next,
\begin{equation*}
\int_0^{C/(2\varepsilon)}dt\,\frac{2\pi\cosh(\varepsilon t)}{\pi^2+4t^2}
\to\int_0^\infty dt\,\frac{2\pi}{\pi^2+4t^2}=\frac\pi2 \tag{6}
\end{equation*}
and
\begin{equation*}
\int_0^{C/(2\varepsilon)}dt\,\frac{4i t \sinh(\varepsilon t)}{\pi^2+4t^2}
=\int_0^{C/2}dz\,\frac{4i z \sinh z}{\varepsilon^2\pi^2+4z^2}
\to\int_0^{A/2}dz\,\frac{4i \sinh z}{4z}=i\,\text{shi}\frac A2, \tag{7}
\end{equation*}
where $\text{shi}$ is the hyperbolic sine integral function.
Collecting the pieces (1)(7), we have
\begin{equation*}
I\to e^{A}(\text{Ei}(A)\text{Ei}(A/2))e^{A}\text{E}_1(A/2)2i e^{A} (\tfrac\pi2+i\,\text{shi}\tfrac A2)
=e^{A}(\text{Ei}(A)i\pi),
\end{equation*}
which is the same expression as the one obtained by Carlo Beenakker.

1$\begingroup$ very interesting! I note that the function you call ${\rm Ei}$ is what Wikipedia calls ${\rm E}_1$, the relation is ${\rm E}_1(x)={\rm Ei}(x)$ for $x>0$. Your final expression can then be simplified to $I=1/A+e^{A}[i\pi+{\rm Ei}\,(A/2)]$. The imaginary part agrees with what I found, but the real part is different: I have ${\rm Re}\,I=e^{A}{\rm Ei}\,(A)$. For $A=1$ you have ${\rm Re}\,I=1.167$ while I get ${\rm Re}\,I=0.697$. $\endgroup$ Aug 7, 2017 at 8:34

$\begingroup$ @CarloBeenakker : Thank you for your comment. I have followed your suggestions. $\endgroup$ Aug 7, 2017 at 13:49

$\begingroup$ Thank you both Iosif and Carlo for your detailed input. I am sorting out what you wrote $\endgroup$ Aug 7, 2017 at 13:57

2$\begingroup$ I made a mistake in estimating $J_2$ and hence in estimating the integral in (2). Now that the mistake is fixed, I get the same expression as Carlo Beenakker's. $\endgroup$ Aug 7, 2017 at 20:42
$$I(\epsilon,B)=\int_0^{\infty}\frac{e^{x}1}{x^{2\epsilon}}\frac{1}{Bi\pi/2\log x}\,dx$$ $$\text{with}\;\;\epsilon=\frac{A}{\log b5/6}\in(0,1)\;\;\text{and}\;\;B=\log b<0$$
The limit $b\rightarrow 0^+$ corresponds to $\epsilon\rightarrow 0^+$, $B\rightarrow\infty$ while the product $\epsilon B\rightarrowA<0$ remains finite. That limit is governed by the small$x$ behaviour of the integrand in $I(\epsilon,B)$, so we may expand $e^{x}1\rightarrowx+{\cal O}(x^2)$ and calculate $$\lim_{\epsilon\rightarrow 0^+}I(\epsilon,A/\epsilon)=\lim_{\epsilon\rightarrow 0^+}\int_0^{1}x^{\epsilon1}\frac{1}{A/\epsilon+\log x+i\pi/2}\,dx=e^{A}\left[i\pi+\text{Ei}\,(A)\right],$$ with ${\rm Ei}(x)=\int_{x}^\infty t^{1}e^{t}\,dt$ the exponential integral function.
The curve below (gold) shows ${\rm Re}\,I$ as a function of $b$ for $A=1$; I was not able to push the numerical integration to smaller $b$, but the numerics is not inconsistent with the analytical result for the asymptote at $e^{1}\,{\rm Ei}\,(1)=0.697$ (blue horizontal line).

$\begingroup$ Numeric calculations do not confirm the divergence in the case $A=1$.. $\endgroup$ Aug 5, 2017 at 17:04

$\begingroup$ Could you kindly elaborate your answer because numeric calculations do not confirm the divergence as $b \to 0+$? $\endgroup$ Aug 5, 2017 at 18:02

$\begingroup$ Maple says that $e^{1}(i\pi+Ei(1))=0.6971748832 1.155727350\,i$. This contradicts the numeric calculations from my answer. $\endgroup$ Aug 5, 2017 at 18:26

$\begingroup$ Maple says that $e^{2}(i\pi+\text{Ei}(2))=0.6704827096 0.4251683315\,i$. This also is not in accordance with the numeric calculations from my answer. I'd like to repeat my want about elaboration of your answer. $\endgroup$ Aug 5, 2017 at 18:35

1$\begingroup$ @CarloBeenaker This was a tough call but I accepted Iosif's answer because I believe that his careful derivation deserves more credit than the 2 upvotes he has so far. In any case, thank you! $\endgroup$ Aug 8, 2017 at 11:12
The math experiment done with Maple suggests there exists a limit (of course, depending on $A$) of the integral as $b \to 0+$:
seq(int(eval((exp(x)1)/(x^(2+1/(ln(b)5/6))*(ln(b/x)I*Pi*(1/2))), b = c),
x = 0 .. infinity, numeric, epsilon = 0.1e2), c = [0.1e3, 0.1e2, 0.1e1, 0.5e1, .1]);
$$ 0.3228143386 1.088234752\,i, 0.3478772540 1.118739511\,i, 0.4172578641 1.160686228\,i, 0.4902769185 1.178652574\,i, 0.5359078669 1.181083764\,i. $$
Addition. In the case $A=2$ one obtains
restart; seq(int(eval((exp(x)1)/(x^(2+2/(ln(b)5/6))*(ln(b/x)I*Pi*(1/2))), b = c),
x = 0 .. infinity, numeric, epsilon = 0.1e1), c = [0.1e3, 0.1e2, 0.1e1, 0.5e1]);
$$ .6100903815.5346046841i, .5958857255.5610607673i, .5824902719.6031233673i, .6107114031.6594924349i $$
Addition 2. Here is the Mathematica result in the case $A=1,\,b=0.0001$:
NIntegrate[((Exp[x]  1)/(x^(2 + 1/(Log[b]  5/6))*(Log[b/x]  I*Pi/2))) /.
b > 0.0001, {x, 0, Infinity}, AccuracyGoal > 3]
$$ 0.535836\, 1.18149 i $$
This is in accordance with the above Maple result $ 0.5359078669 1.181083764\,i$.
Also
NIntegrate[((Exp[x]  1)/(x^(2 + 1/(Log[b]  5/6))*(Log[b/x]  I*Pi/2))) /.
b > 0.000001, {x, 0, Infinity}, AccuracyGoal > 2]
$0.617192  1.17718 i. $

$\begingroup$ Remark. The integral is calculated for $A=1, b \in \{0.0001, 0.001, 0.01, 0.05\}$ with the relative error $0.001$ and for $A=2, b \in \{ 0.0001, 0.001,0,01 ,0.05\}$ with the relative error $0.01$. $\endgroup$ Aug 5, 2017 at 17:55

$\begingroup$ notice that $b$ appears as $\log b$, so the convergence in $b$ is very slow, even for $b=0.0001$ you only have $1/\log b\approx 0.1$, so not very small. $\endgroup$ Aug 6, 2017 at 11:11

$\begingroup$ Coud you kindly base your statement? No remainder estimate is given in your answer. The limit is taken over $b$, not over $\log b$. $\endgroup$ Aug 6, 2017 at 11:21

$\begingroup$ I prefer arguments over unbased sentences e.g. "so it is reasonable to assume ". How about you? $\endgroup$ Aug 6, 2017 at 11:26

$\begingroup$ It seems you are right because Mathematica outputs $ 0.686336\, 1.17138 i$ for $A=1, b=0.00000001$. $\endgroup$ Aug 6, 2017 at 11:35