Leading term of an integral for $a\to +\infty$

Question

Given $\int_{\frac{1}{b}}^{a}\frac{1}{(1+a-x)^{\rho}(1+bx)}dx$, with $a, b>0$ and $\rho>2$, I want to find the leading term of the expansion for $a\to +\infty$. I followed the procedure showed in an answer to a similar question I asked recently (Asymptotics of an integral depending on a parameter) which led me to $\frac{\ln (ab)}{a^{\rho}b}$. Then, to verify my answer, I tried to compare the result with calculation of the integral for large $a$ made with Wolfram Alpha. Sadly, my result seems to go too fast to zero to be the right one. Can someone tell me if my result is correct or not? Thanks

Answer 1

First claim that the leading term is $\color{blue}{\frac{1}{b(\rho-1)a}}$.

To see this, it's important to note that \begin{align*} I& :=\int_{\frac{1}{b}}^{a}\frac{\mathrm{d}x}{\left(1+a-x\right)^{\rho}(1+bx)}\\ & =\int_{\frac{1}{b}}^{a}\frac{\mathrm{d}x}{\left(1+a-x\right)^{\rho}bx}\\ & \ \ \ -\int_{\frac{1}{b}}^{a}\frac{\mathrm{d}x}{\left(1+a-x\right)^{\rho}bx(1+bx)}\\ & =:I_1-I_2. \end{align*} Integration by parts can get \begin{align*} I_1& =\frac{1}{b(\rho-1)}\int_{b^{-1}}^a\frac{\mathrm{d}\left(1+a-x\right)^{1-\rho}}{x}\\ & =\frac{1}{b(\rho-1)a}-\frac{\left(1+a-b^{-1}\right)^{1-\rho}}{\rho-1}\\ & \ \ \ +\frac{1}{b(\rho-1)}\int_{b^{-1}}^a\frac{\mathrm{d}x}{x^2\left(1+a-x\right)^{\rho-1}}. \end{align*} For $r>1$, and fixed $\delta\in \left(\frac{1}{b(a+1)},\frac{a}{a+1}\right)$, we have \begin{align*} J_1& :=\int_{\frac{1}{b(a+1)}}^{\delta}\frac{\mathrm{d}t}{t^2\left(1-t\right)^r}\sim \int_{\frac{1}{b(a+1)}}^{\delta}\frac{\mathrm{d}t}{t^2}\sim ba;\\ J_2& :=\int_{\delta}^{1-\delta}\frac{\mathrm{d}t}{t^2\left(1-t\right)^r}\leqslant K_{\delta}<+\infty;\\ J_3& :=\int_{1-\delta}^{\frac{a}{a+1}}\frac{\mathrm{d}t}{t^2\left(1-t\right)^r}\sim \int_{1-\delta}^{\frac{a}{a+1}}\frac{\mathrm{d}t}{\left(1-t\right)^r}\sim \frac{a^{r-1}}{r-1} \end{align*} hold as $a\to +\infty$.

Substituting $x=(a+1)t$ can get \begin{align*} J(r)& :=\int_{b^{-1}}^a\frac{\mathrm{d}x}{x^2\left(1+a-x\right)^r}\\ & =\frac{1}{\left(a+1\right)^{r+1}}\int_{\frac{1}{b(a+1)}}^{\frac{a}{a+1}}\frac{\mathrm{d}t}{t^2\left(1-t\right)^r}\\ & =\frac{J_1+J_2+J_3}{\left(a+1\right)^{r+1}}=\mathcal{O}\left(a^{\max\{-r,-2\}}\right). \end{align*} Thus \begin{align*} I_1& =\frac{1}{b(\rho-1)a}+\mathcal{O}\left(a^{1-\rho}\right)+\frac{J(\rho-1)}{b(\rho-1)}\\ & =\frac{1}{b(\rho-1)a}+\mathcal{O}\left(a^{\max\{1-\rho,-2\}}\right);\\ I_2& \leqslant b^{-2}J(r)=\mathcal{O}\left(a^{-2}\right). \end{align*} Hence as $a\to +\infty$, $$\color{blue}{I=\frac{1}{b(\rho-1)a}+\mathcal{O}\left(a^{\max\{1-\rho,-2\}}\right).}$$

Answer 2

One way to tackle the asymptotics of this integral is to make a change of variables: $$x=\Big(\frac{1}{b}+a+1\Big)t-\frac{1}{b}$$

The point of this change of variables is to remove all dependence of the integrand on $a$ and transfer it to the boundary, so one can take derivatives more easily.

Applying this change of variables we get

$$I=\int_{1/b}^{a}\frac{dx}{(1+a-x)^{\rho}(1+bx)}=\frac{1}{(a+1+\frac{1}{b})^{\rho}}\int_{\frac{2}{1+b+ab}}^{\frac{ab+1}{ab+b+1}}\frac{dx}{x(1-x)^{\rho}}\equiv\frac{1}{b(a+1+\frac{1}{b})^{\rho}}J$$

and we take a derivative with respect to $a$ we obtain

$$\frac{dJ}{da}=\frac{(a+\frac{1+b}{b})^{\rho-2}}{1-\frac{1}{a+\frac{1+b}{b}}}+\frac{1}{(a+\frac{1+b}{b})(1-\frac{2}{1+b+ab})^{\rho}}=a^{\rho-2}+\mathcal{O}(a^{\rho-3})%\begin{Bmatrix}\mathcal{O}(1/a)& 2<\rho<3\\\mathcal{O}(a^{\rho-3})& \rho>3\end{Bmatrix}%$$

from which we conclude that

$$J=\frac{a^{\rho-1}}{\rho-1}+\mathcal{O}(a^{\rho-2})$$

and finally

$$I=\frac{1}{b(\rho-1)a}+\mathcal{O}(a^{-2})$$

Note: This method can be extended to yield subleading corrections to this integral. Also, this integral contains $\frac{\ln a}{a^\rho}$ as a subleading correction, but to leading order, it is overshadowed by other corrections that decay slower.