Integral of $\int_{r_{0}}^{r_{1}}\frac{1}{x^{2}}e^{-a*arctan(x/b)}dx$

Question

I'm trying to solve this integral for $r_{0}\rightarrow 0$, $r_{1}\rightarrow +\infty$ with $a$ and $b$ positive numbers but I don't even know where to start. Looking for a closed-form solution if possible.

Answer 1

$$\int_{0}^{\infty}\dfrac{1}{x^{2}}e^{-a\cdot\arctan(x/b)}dx\to\infty$$ Since $$e^{-a\cdot\arctan(x/b)}\sim 1-\frac{ax}{b}\text{ for }x\to 0$$ So you are trying to integrate a function that scales like $\dfrac{1}{x^2}$ for $x\to 0$
But it deverges.
I suggest you change the integral to something like $$\int_{1}^{\infty}\dfrac{1}{x^{2}}e^{-a\cdot\arctan(x/b)}dx$$

Answer 2

$$I=\int\frac{1}{x^2}e^{-a \tan ^{-1}\left(\frac{x}{b}\right)}\,dx$$ Let $x=b \,t$ at least to get rid of one constant $$J=\frac 1 b \int \frac{e^{-a \tan ^{-1}(t)}}{t^2}\, dt$$ Using the complex representation of the arctangent function, this leads to a quite nasty Gaussian hypergeometric function. $$J=-\frac {a-2i}{1+\frac 14 a^2}\left(\frac{i+t}{i-t}\right)^{1-\frac{i a}{2}}\, _2F_1\left(2,\frac{2-i a}{2};\frac{4-i a}{2};\frac{i+t}{i-t}\right)$$ which does not make any problem

Writing $$e^{-a \tan ^{-1}(t)}=\sum_{n=0}^\infty \alpha_n\,t^n$$ with $$\alpha_0=1 \qquad \alpha_1=-a\qquad \alpha_n=-\frac{a\,\alpha_{n-1}+(n-2)\,\alpha_{n-2} } {n}$$