Asymptotics of an integral depending on a parameter

Question

Given the integral $\int_{1}^{+\infty}\frac{1}{(1+a^2+x^2)^{\frac{\rho}{2}}(1+2\sqrt{E}x)}dx$, with $a,E>0$ and $\rho>2$, I want to know how to determine how it goes to 0 as $a\to +\infty$. Maybe the answer is trivial, but I prefer being cautious

Answer 1

I'll write $b=2\sqrt E$ to simplify things.

To get a systematic expansion, write

\begin{eqnarray} \int_1^a\frac1{(1+a^2+x^2)^\frac\rho2(1+bx)}\,\mathrm dx &=& \frac1{a^\rho b}\int_1^a\frac1{(1+\frac{x^2+1}{a^2})^\frac\rho2(1+\frac1{bx})x}\,\mathrm dx \\ &=& \frac1{a^\rho b}\int_1^a\sum_{j=0}^\infty\sum_{k=0}^\infty\binom{-\frac\rho 2}j\left(\frac{x^2+1}{a^2}\right)^j\left(-\frac1{bx}\right)^k\frac1x\,\mathrm dx \\ &=& \sum_{j=0}^\infty\sum_{k=0}^\infty\sum_{n=0}^j\frac{(-1)^k}{a^{\rho+2j}b^{k+1}}\binom{-\frac\rho 2}j\binom jn\int_1^ax^{2n-k-1}\,\mathrm dx\;. \\ &=& \sum_{j=0}^\infty\sum_{k=0}^\infty\sum_{n=0}^j\frac{(-1)^k}{a^{\rho+2j}b^{k+1}}\binom{-\frac\rho 2}j\binom jn \begin{cases} \ln a&2n-k=0\;,\\ \frac{a^{2n-k}-1}{2n-k}&\text{otherwise}\;. \end{cases} \end{eqnarray}

The leading term is the one for $j=k=0$, which yields

$$ \frac{\ln a}{a^\rho b}\;, $$

as in @stokes-line’s answer. The next term comes from contributions for $k=0$, $n=j\gt0$:

$$ \sum_{j=1}^\infty\frac1{a^{\rho+2j}b}\binom{-\frac\rho 2}j\frac{a^{2j}}{2j} = -\frac1{a^\rho b}\frac\rho4\,_3F_2\left(1,1,\frac\rho2+1;2,2;-1\right)\;, $$

where $_3F_2$ is a generalized hypergeometric series, and for $j=0$, $k\gt0$:

$$ \sum_{k=1}^\infty\frac{(-1)^k}{a^\rho b^{k+1}}\frac{-1}{-k} = -\frac{\ln\left(1+\frac1b\right)}{a^\rho b}\;. $$

The term after that comes from $k=1$, $n=j$:

$$ -\sum_{j=0}^\infty\frac1{a^{\rho+2j}b^2}\binom{-\frac\rho2} j\frac{a^{2j-1}}{2j-1} = \frac1{a^{\rho+1}b^2}\,_2F_1\left(-\frac12,\frac\rho2;\frac12;-1\right)\;. $$

For the part beyond $a$, we need to expand in $\frac1x$ instead, yielding

\begin{eqnarray} \int_a^\infty\frac1{(1+a^2+x^2)^\frac\rho2(1+bx)}\,\mathrm dx &=& \int_a^\infty\frac1{bx^{\rho+1}}\frac1{\left(1+\frac{a^2+1}{x^2}\right)^\frac\rho2\left(1+\frac1{bx}\right)}\,\mathrm dx \\ &=& \int_a^\infty\frac1{bx^{\rho+1}}\sum_{j=0}^\infty\sum_{k=0}^\infty\binom{-\frac\rho2}j\left(\frac{a^2+1}{x^2}\right)^j\left(-\frac1{bx}\right)^k\,\mathrm dx \\ &=& \sum_{j=0}^\infty\sum_{k=0}^\infty\binom{-\frac\rho2}j\frac{(-1)^k}{a^{\rho+2j+k}b^{k+1}}(a^2+1)^j\int_1^\infty t^{-(\rho+2j+k+1)}\,\mathrm dt \\ &=& \sum_{j=0}^\infty\sum_{k=0}^\infty\sum_{n=0}^j\binom{-\frac\rho2}j\binom jn\frac{(-1)^k}{a^{\rho+2j+k-2n}b^{k+1}}\frac1{\rho+2j+k}\;. \end{eqnarray}

Here there is no logarithmic term; the first term comes from $k=0$, $n=j$:

$$ \sum_{j=0}^\infty\binom{-\frac\rho2}j\frac1{a^\rho b}\frac1{\rho+2j}=\frac1{a^\rho b}\frac1\rho\,_2F_1\left(\frac\rho2,\frac\rho2;\frac\rho2+1;-1\right)\;. $$

The next term comes from $k=1$, $n=j$:

$$ -\sum_{j=0}^\infty\binom{-\frac\rho2}j\frac1{a^{\rho+1}b^2}\frac1{\rho+2j+1}=-\frac1{a^{\rho+1}b^2}\frac1{\rho+1}\,_2F_1\left(\frac\rho2,\frac{\rho+1}2;\frac{\rho+3}2;-1\right)\;. $$

Thus, altogether we have the expansion

$$ \int_1^a\frac1{(1+a^2+x^2)^\frac\rho2(1+bx)}\,\mathrm dx \\ = \frac{\ln a-\ln\left(1+\frac1b\right)}{a^\rho b} + \left(\frac1\rho\,_2F_1\left(\frac\rho2,\frac\rho2;\frac\rho2+1;-1\right)-\frac\rho4\,_3F_2\left(1,1,\frac\rho2+1;2,2;-1\right)\right)\frac1{a^\rho b} \\ +\left(_2F_1\left(-\frac12,\frac\rho2;\frac12;-1\right)-\frac1{\rho+1}\,_2F_1\left(\frac\rho2,\frac{\rho+1}2;\frac{\rho+3}2;-1\right)\right)\frac1{a^{\rho+1}b^2} \\ +O\left(\frac1{a^{\rho+2}}\right)\;. $$

For integer values of $\rho$, the hypergeometric series yield simple expressions. For instance, for $\rho=2$ the expansion takes the form

$$ \int_1^a\frac1{(1+a^2+x^2)(1+bx)}\,\mathrm dx \\ = \frac{\ln a-\ln\left(1+\frac1b\right)}{a^2 b} + \left(\frac12\,_2F_1\left(1,1;2;-1\right)-\frac12\,_3F_2\left(1,1,2;2,2;-1\right)\right)\frac1{a^2 b} \\ +\left(_2F_1\left(-\frac12,1;\frac12;-1\right)-\frac13\,_2F_1\left(2,\frac32;\frac52;-1\right)\right)\frac1{a^3b^2} \\ +O\left(\frac1{a^4}\right)\;. \\ = \frac{\ln a-\ln\left(1+\frac1b\right)}{a^2 b} +\frac{\frac54+\frac\pi8}{a^3b^2} +O\left(\frac1{a^4}\right)\;. $$

The terms proportional to $\frac1{a^2b}$ cancel. For $\rho=3$, the result is

$$ \int_1^a\frac1{(1+a^2+x^2)^\frac32(1+bx)}\,\mathrm dx \\ = \frac{\ln a-\ln\left(1+\frac1b\right)}{a^3 b} + \left(\frac13\,_2F_1\left(\frac32,\frac32;\frac52;-1\right)-\frac34\,_3F_2\left(1,1,\frac52;2,2;-1\right)\right)\frac1{a^3 b} \\ +\left(_2F_1\left(-\frac12,\frac32;\frac12;-1\right)-\frac14\,_2F_1\left(\frac32,2;3;-1\right)\right)\frac1{a^4b^2} \\ +O\left(\frac1{a^5}\right)\;. \\ = \frac{\ln a-\ln\left(1+\frac1b\right)}{a^3 b} +\frac{\ln2-1}{a^3 b} +\frac2{a^4b^2} +O\left(\frac1{a^5}\right)\;. $$

For $\rho=4$, the result is

$$ \int_1^a\frac1{(1+a^2+x^2)^2(1+bx)}\,\mathrm dx \\ = \frac{\ln a-\ln\left(1+\frac1b\right)}{a^4 b} + \left(\frac14\,_2F_1\left(2,2;3;-1\right)-\,_3F_2\left(1,1,3;2,2;-1\right)\right)\frac1{a^4 b} \\ +\left(_2F_1\left(-\frac12,2;\frac12;-1\right)-\frac15\,_2F_1\left(2,\frac52;\frac72;-1\right)\right)\frac1{a^5b^2} \\ +O\left(\frac1{a^6}\right)\;. \\ = \frac{\ln a-\ln\left(1+\frac1b\right)}{a^4 b} -{\frac12}\frac1{a^4 b} +\frac{3\pi}4\frac1{a^5b^2} +O\left(\frac1{a^6}\right)\;. $$

For $\rho=5$, the result is

$$ \int_1^a\frac1{(1+a^2+x^2)^\frac52(1+bx)}\,\mathrm dx \\ = \frac{\ln a-\ln\left(1+\frac1b\right)}{a^5 b} + \left(\frac14\,_2F_1\left(\frac52,\frac52;\frac72;-1\right)-\,_3F_2\left(1,1,\frac72;2,2;-1\right)\right)\frac1{a^5 b} \\ +\left(_2F_1\left(-\frac12,\frac52;\frac12;-1\right)-\frac15\,_2F_1\left(\frac52,3;4;-1\right)\right)\frac1{a^6b^2} \\ +O\left(\frac1{a^7}\right)\;. \\ = \frac{\ln a-\ln\left(1+\frac1b\right)}{a^5 b} +\frac{\ln2-\frac43}{a^5 b} +\frac83\frac1{a^6b^2} +O\left(\frac1{a^7}\right)\;. $$

Answer 2

The leading asymptotics can be found realizing that at large $a$ the integral comes from $1\lesssim x\lesssim a$ and is logarithmic in this range. As a result: $$ I\sim \frac{\ln a}{2 \sqrt{E} a^\rho}. $$