Tighter asymptotic upper bound for exp integral

Question

I'm interested in the asymptotic behavior of the function $$ f(x) = \int_{0}^{\infty}\frac{e^{-t}dt}{(x+t)^2},\;\;x>0, $$ as $x\to0+$. The integral diverges at $x=0$. I would like to know how how fast. One obvious idea is to argue that $e^{-t}\le 1$ for $t\ge0$ and immediately deduce that $f(x)\le \frac {1}{x}$. How would I go about tightening this bound? The function can be expressed as the difference $\frac{1}{x}-e^{x}E_1(x)$, where $E_1(x)$ denotes the exponential integral.

Answer 1

Herein, we develop a full expansion of the function $\displaystyle f(x)=\int_0^\infty\frac{e^{-t}}{(x+t)^2}\,dt$.

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Let $f(x)$ be given by the integral

$$\begin{align} f(x)&=\int_{0}^{\infty}\frac{e^{-t}dt}{(x+t)^2}\,dt\\\\ &=e^x\int_x^\infty\frac{e^{-t}}{t^2}\,dt\tag1 \end{align}$$

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Next, we Integrate by parts the integral on the right-hand side of $(1)$ with $u=e^{-t}$ and $v=-1/t$ to obtain

$$\int_x^\infty\frac{e^{-t}}{t^2}\,dt=\frac{e^{-x}}{x}-\int_x^\infty \frac{e^{-t}}{t}\,dt$$

whereby $f(x)$ becomes

$$f(x)=\frac1x-e^x\int_x^\infty \frac{e^{-t}}{t}\,dt \tag 2$$

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Then, integrating by parts the integral on the right-hand side of $(2)$ with $u=e^{-t}$ and $v=\log(t)$ yields

$$\begin{align} \int_x^\infty \frac{e^{-t}}{t}\,dt&=-e^{-x}\log(x)+\int_x^\infty e^{-t}\log(t)\,dt\\\\ &=-e^{-x}\log(x)-\gamma -\int_0^x e^{-t}\log(t)\,dt \end{align}$$

whereby $f(x)$ becomes

$$\begin{align} f(x)&=\frac1x+\log(x)+\gamma \underbrace{e^x}_{=1+o(1)}+e^x\underbrace{\int_0^x e^{-t}\log(t)\,dt}_{=o(1)}\tag 3 \\\\ &=\frac1x+\log(x)+\gamma +o(1) \end{align}$$

which provides an expansion up to terms that are $o(1)$ as $x\to 0$ (i.e., terms that approach $0$ as $x\to 0$).

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DEVELOPING THE FULL ASYMPTOTIC SERIES:

In order to develop the full asymptotic series, we need to expand $I(x)=\int_0^xe^{-t}\log(t)\,dt$.

Using $e^{-t}=\sum_{n=0}^\infty\frac{(-1)^n\,t^n}{n!}$ reveals

$$\begin{align} I(x)&=\int_0^x e^{-t}\log(t)\,dt\\\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^x t^n\log(t)\,dt\\\\ &=\log(x)\sum_{n=0}^\infty \frac{(-1)^nx^{n+1}}{(n+1)!} -\sum_{n=0}^\infty\frac{(-1)^{n}x^{n+1}}{n!\,(n+1)^2}\\\\ &=-\log(x)\underbrace{\sum_{n=1}^\infty\frac{(-1)^nx^n}{n!}}_{=e^{-x}-1}+\sum_{n=1}^\infty\frac{(-1)^{n}x^{n}}{n\,n!}\tag 4 \end{align}$$

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Next, we expand $e^xI(x)$. Using $(4)$ reveals that

$$\begin{align} e^xI(x)&=-\log(x)+e^x\log(x)+e^x\sum_{n=1}^\infty\frac{(-1)^{n}x^{n}}{n\,n!}\\\\ &=\log(x)\sum_{n=1}^\infty\frac{x^n}{n!}+e^x\sum_{n=1}^\infty\frac{(-1)^{n}x^{n}}{n\,n!}\\\\ &=\log(x)\sum_{n=1}^\infty\frac{x^n}{n!}+\sum_{m=0}^\infty\frac{x^m}{m!}\sum_{n=1}^\infty\frac{(-1)^{n}x^{n}}{n\,n!}\\\\ &=\log(x)\sum_{n=1}^\infty\frac{x^n}{n!}+\sum_{n=1}^\infty \left(\sum_{p=1}^n\frac{(-1)^p}{p}\binom{n}{p}\right) \frac{x^n}{n!}\tag5 \end{align}$$

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Finally, using $(5)$ in $(3)$ yields

$$\begin{align} \bbox[5px,border:2px solid #C0A000]{f(x)=\frac1x +\log(x)+\gamma +\log(x)\sum_{n=1}^\infty \frac{x^n}{n!}+\sum_{n=1}^\infty \left(\gamma+\sum_{p=1}^n \frac{(-1)^p}{p}\binom{n}{p}\right)\frac{x^n}{n!}} \end{align}$$

which is the full asymptotic expansion of $f(x)$.

Answer 2

We can use the method of integration by parts to obtain the desired asymptotic behaviour: $\int_{0}^{\infty}\frac{e^{-t}}{(x+t)^2}dt =-\int_{0}^{\infty}e^{-t}d(\frac{1}{x+t}) =\frac{1}{x}-\int_{0}^{\infty}\frac{e^{-t}}{x+t}dt =\frac{1}{x}+\log{x}-\int_0^{\infty}\log{(x+t)}e^{-t}dt$. Since the last integral is convergent when $x\to0$, we can use the dominated convergence theorem to get the limit.

Answer 3

Integrating by parts, raising $dt/(x + t)^2 \mapsto -1/(x+t)$ and lowering $e^{-t}\mapsto -dt~e^{-t}$ gives $$f(x) = \frac 1x - \int_0^\infty dt~\frac{e^{-t}}{t + x} = \frac 1 x-\int_x^\infty du~u^{-1}e^{-u+x}=\frac 1x - e^x ~\Gamma(0, x),$$where $\Gamma$ is the upper incomplete gamma function. Per Wikipedia, this has a series expansion, $$f(x) = \frac 1x + e^x \left(\gamma + \ln x + \sum_{k=1}^\infty \frac{(-x)^k}{k~k!}\right),$$ therefore the divergence at $x=0$ behaves asymptotically like $\frac 1x + \ln x.$