Asymptotic expansion of $I(\alpha;\epsilon) = \int_0^{\infty}\frac{dx}{(\epsilon^2+x^2)^{\alpha/2}(1+x)}$

Question

Question: Evaluate the first two terms of as $\epsilon \to 0$ of $$I(\alpha;\epsilon) = \int_0^{\infty}\frac{dx}{(\epsilon^2+x^2)^{\alpha/2}(1+x)},$$ for $\alpha = \frac{1}{2}, 1, 2$, if $$C(\alpha) = \int_0^{\infty}\left[\frac{1}{(1+u^2)^{\alpha/2}} - \frac{1}{(1+u)^{\alpha}}\right]du.$$

My approach: The integral can be rewritten as $$I(\alpha;\epsilon) = \int_0^{\infty}\frac{x^{-\alpha}}{\left[(\epsilon/x)^2 +1)^{\alpha/2}(1+x)\right]}dx.$$ The integrand can be Taylor expanded only when $(\epsilon/x)^2 \ll 1$, that is, when $\epsilon \ll x$. I am not sure how to systematically proceed after this. Can someone please help? Thanks in advance.

Answer 1

Making use of Mathematica, here these are:

AsymptoticIntegrate[1/(\[Epsilon]^2 + x^2)^(1/2)/(1 + x), {x, 0, Infinity}, {\[Epsilon],  0, 2}, Assumptions -> \[Epsilon] > 0]

$$\frac{1}{4} \epsilon ^2 \left(2 \log (\epsilon )+\gamma -1+2 \psi ^{(0)}\left(\frac{3}{2}\right)-\psi ^{(0)}\left(-\frac{1}{2}\right)\right)+\epsilon +\frac{1}{2} \left(-2 \log (\epsilon )-\gamma -\psi ^{(0)}\left(\frac{1}{2}\right)\right)$$

AsymptoticIntegrate[1/(\[Epsilon]^2 + x^2)/(1 + x), {x, 0, Infinity}, {\[Epsilon], 0, 2}]

$$\frac{1}{2} \pi \sqrt{\frac{1}{\epsilon ^2}}-\frac{\pi }{2 \sqrt{\frac{1}{\epsilon ^2}}}-\frac{1}{2} \epsilon ^2 \log \left(\epsilon ^2\right)+\frac{\log \left(\epsilon ^2\right)}{2} $$

AsymptoticIntegrate[1/(\[Epsilon]^2 + x^2)^(1/4)/(1 + x), {x, 0, Infinity}, {\[Epsilon], 0,  2}]

$$\frac{2 \epsilon ^{3/2}}{3}-\frac{16 \sqrt{\pi } \epsilon ^{5/2} \Gamma \left(\frac{3}{4}\right)}{15 \Gamma \left(-\frac{3}{4}\right)}-\frac{\pi \epsilon ^2}{4}+\frac{8 \sqrt{\pi } \sqrt{\epsilon } \Gamma \left(\frac{3}{4}\right)}{3 \Gamma \left(-\frac{3}{4}\right)}+\pi $$

Answer 2

$$I = \int\frac{dx}{(\epsilon^2+x^2)^{\frac\alpha 2}(1+x)}$$

Assuming that $(x,\alpha,\epsilon)$ are all positive, Mathematica returns $$I=\frac 1 {\alpha (x+1)^{\alpha } }\,\,F_1\left(\alpha ;\frac{\alpha }{2},\frac{\alpha }{2};\alpha +1;\frac{1-i \epsilon }{x+1},\frac{1+i \epsilon }{x+1}\right)$$ where $F_1$ is the Appell hypergeometric function of two variables.

The definite integral is

$$\frac 1 \alpha\,\, F_1\left(\alpha ;\frac{\alpha }{2},\frac{\alpha }{2};\alpha +1;1-i \epsilon ,1+i \epsilon \right)$$ which is not the most pleasant to expand as a series (but doable). What is difficult is to have the expansion for general $\alpha$.

I shall not repeat the remaining but the asymptotics @user64494 gave in his/her answer are really good.