Integral asymptotic expansion

Question

How can the following identity be proven: $$ \int_0^\infty dy \frac{y}{(1+y)^2}e^{-\alpha y} = \ln{(1/\alpha)}+\mathcal{O}(1), $$ for $\alpha\ll1$?

Answer 1

Too long for a comment $$\int_0^\infty \frac{y}{(1+y)^2}e^{-\alpha y}dy\overset{t=\alpha y}{=}\int_0^\infty \frac{t}{(\alpha+t)^2}e^{-t}dt=\int_0^\infty \frac{e^{-t}}{\alpha+t}dt-\alpha\int_0^\infty \frac{e^{-t}}{(\alpha+t)^2}dt$$ $$\overset{IBP}{=}\int_0^\infty \frac{e^{-t}}{\alpha+t}dt+\alpha\frac{e^{-t}}{\alpha+t}\,\bigg|_{t=0}^\infty+\alpha\int_0^\infty \frac{e^{-t}}{\alpha+t}dt$$ $$=-1+(1+\alpha)\int_0^\infty \frac{e^{-t}}{\alpha+t}dt$$ $$\overset{IBP}{=}-1+(1+\alpha)\,e^{-t}\ln(\alpha+t)\,\bigg|_{t=0}^\infty+(1+\alpha)\int_0^\infty\ln(\alpha+t)\,e^{-t}dt$$ $$=-1-(1+\alpha)\ln\alpha+(1+\alpha)e^\alpha\int_\alpha^\infty\ln t\,e^{-t}dt$$ $$=-1-(1+\alpha)\ln\alpha+(1+\alpha)e^\alpha\int_0^\infty\ln t\,e^{-t}dt-(1+\alpha)e^\alpha\int_0^\alpha\ln t\,e^{-t}dt$$ $$=-1-(1+\alpha)\ln\alpha-(1+\alpha)e^\alpha\gamma-(1+\alpha)e^\alpha\int_0^\alpha\ln t\left(1-t+\frac{t^2}2-\frac{t^3}{3!}+...\right)dt$$ $$=-\ln\alpha-1-\gamma-2\alpha\ln\alpha+(1-2\gamma)\alpha+O(\alpha^2\ln\alpha)$$

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Numeric check with WA $$\text {at}\,\alpha=0.01\quad\int_0^\infty \frac{y}{(1+y)^2}e^{-0.01 y}dy=3.1193...\quad\text{approximation}=3.11851...$$

Answer 2

$$\int \frac{y}{(1+y)^2}\,e^{-\alpha y}\,dy=\frac{e^{-\alpha y}}{y+1}+(\alpha +1)\, e^{\alpha }\, \text{Ei}(-\alpha (y+1) )$$

Assuming $\alpha >0$

$$I=\int_0^\infty \frac{y}{(1+y)^2}\,e^{-\alpha y}\,dy=-1-(\alpha +1) \,e^{\alpha }\, \text{Ei}(-\alpha )$$

Expanding as Taylor series around $\alpha=0$ $$I=\sum_{n=0}^\infty P_n\,\alpha ^n$$ where the $P_n$ write $$P_n=a_n - b_n \,L \qquad \text{with} \qquad L=\log(\alpha)+\gamma$$

The first coefficients are $$\left( \begin{array}{ccc} n & a_n & b_n \\ 0 & -1 & 1 \\ 1 & 1 & 2 \\ 2 & \frac{7}{4} & \frac{3}{2} \\ 3 & \frac{19}{18} & \frac{2}{3} \\ 4 & \frac{113}{288} & \frac{5}{24} \\ 5 & \frac{127}{1200} & \frac{1}{20} \\ 6 & \frac{323}{14400} & \frac{7}{720} \\ \end{array} \right)$$

which can be excellent approximations of $I$ over a "large" range of $\alpha$ if a sufficient number of terms is used.

For $\alpha=0.1$ $$\left( \begin{array}{cc} 0 & 0.725369428 \\ 1 & 1.170443314 \\ 2 & 1.213823855 \\ 3 & 1.216029657 \\ 4 & 1.216104838 \\ 5 & 1.216106759 \\ 6 & 1.216106799 \\ \cdots & \cdots \\ \infty & 1.216106799 \\ \end{array} \right)$$

Answer 3

Another method at arriving at the same solution as @Svyatoslav is by considering the Laplace transform with small parameters which is based on the Mellin transform. Details are provided in the link above. Essentially, one transforms the problem into evaluating an inverse Mellin transform such that the asymptotic terms are given by the residues.

Let

$$ I = \int_0^{\infty} H(y) e^{-\alpha y}dy $$ with

$$ H(y) = \frac{y}{(1+y)^2} \Rightarrow \mathcal{M}[H(y)](z) = z\Gamma(z)\Gamma(1-z) $$

For the function

$$ F(z) = -\alpha^{z-1}\Gamma(1-z)\mathcal{M}[H(y)](z) = -\alpha^{z-1}z\Gamma(z)\Gamma^2(1-z) $$

we calculate the residues at $z=0,1,2,...$.

At $z=0$, the residue is $0$. At $z=1,2,...$ the pole is of order $2$ as seen from $\Gamma^2(1-z) = (1-z)^{-2}\Gamma(2-z)$ (and so on). I.e. the residue at $z=k$ is given by

$$ \text{res}_{z=k}F(z) = \lim_{z\to k} \frac{d}{dz}\left[(z-k)^2F(z) \right] $$

EDIT: I have changed the derivation to calculate the residues at $z=k$ where $k$ is a positive integer. Previously I only did it for $z=1,2,3$.

We may relatively easily find the residue at $z=k$. Writing

\begin{align} \Gamma(1-z)&=\Gamma(k+1-z)\prod_{j=0}^{k-1} (1+j-z)^{-2} \\ \Rightarrow F(z)&=\alpha^{z-1}z\Gamma(z)\Gamma^2(k+1-z)\prod_{j=0}^{k-1}(1+j-z)^{-2} \\ \Rightarrow (z-k)^2F(z) &= \alpha^{z-1}z\Gamma(z)\Gamma^2(k+1-z)\prod_{j=0}^{k-2}(1+j-z)^{-2} \end{align}

taking the derivative wrt $z$ gives

\begin{align} \frac{d}{dz}\left[(z-k)^2F(z) \right] &= \alpha^{z-1}\Gamma(z)\Gamma^2(k+1-z)(z\ln\alpha-2z\psi(k+1-z)+z\psi(z)+1)\prod_{j=0}^{k-2}(1+j-z)^{-2} \\ &+ 2\alpha^{z-1}z\Gamma(z)\Gamma^2(k+1-z)\sum_{l=0}^{k-2}(l+1-z)^{-1}\prod_{j=0}^{k-2}(1+j-z)^{-2} \end{align}

Evaluating as $z\to k$ gives the residue (after some simplification)

\begin{align} \text{res}_{z=k}F(z) &= -\frac{1}{\Gamma(k)}\alpha^{k-1}(k\ln\alpha+2\gamma k+k\psi(k)+1) \\ &+\frac{2}{\Gamma(k)}\alpha^{k-1}k(\gamma+\psi(k)) \\ &= -\frac{1}{\Gamma(k)}\alpha^{k-1}(k\ln\alpha-k\psi(k)+1) \\ \end{align}

E.g. the three first residues are given by

\begin{align} \text{res}_{z=1}F(z) &= -\ln \alpha - \gamma - 1 \\ \text{res}_{z=2}F(z) &= -2\alpha \ln \alpha - 2\gamma \alpha +\alpha \\ \text{res}_{z=3}F(z) &=-\frac{1}{4}\alpha^2 (6\ln\alpha+6\gamma-7) \end{align}

and the full asymptotic expansion is given by

$$ I \sim \sum_{k=1}^{\infty} \frac{\alpha^{k-1}}{\Gamma(k)}(k\psi(k)-k\ln\alpha-1), \quad \alpha \to 0_+ $$

Note that for the first two terms, we obtain the same as Svyatoslav.