I would like to compute the following integral: $$ \int_{-\infty}^{+\infty}\frac{e^{i\sqrt{a^2-x^2}}}{\sqrt{a^2-x^2}}\mathrm{d}x\quad,\quad a\in\mathbb{R}$$ Where I extend the sqrt function to complex number such that $\sqrt{a^2-x^2} = i\sqrt{x^2-a^2}$ when $|x|>|a|$. So I can split the integral in three: $$ \int_{-\infty}^{+\infty}\frac{e^{i\sqrt{a^2-x^2}}}{\sqrt{a^2-x^2}}\mathrm{d}x = \int_{-\infty}^{-a}\frac{e^{-\sqrt{x^2-a^2}}}{i\sqrt{x^2-a^2}}\mathrm{d}x + \int_{a}^{+\infty}\frac{e^{-\sqrt{x^2-a^2}}}{i\sqrt{x^2-a^2}}\mathrm{d}x + \int_{-a}^{a}\frac{e^{i\sqrt{a^2-x^2}}}{\sqrt{a^2-x^2}}\mathrm{d}x$$ And then by parity: $$ \int_{-\infty}^{+\infty}\frac{e^{i\sqrt{a^2-x^2}}}{\sqrt{a^2-x^2}}\mathrm{d}x = -2i \int_{a}^{+\infty}\frac{e^{-\sqrt{x^2-a^2}}}{\sqrt{x^2-a^2}}\mathrm{d}x + 2\int_{0}^{a}\frac{e^{i\sqrt{a^2-x^2}}}{\sqrt{a^2-x^2}}\mathrm{d}x$$ These two integrals are convergent, for $x\rightarrow +\infty$ it is obvious, for $x\rightarrow a$ we have: $$\frac{1}{\sqrt{x^2-a^2}}=\frac{1}{\sqrt{(x-a)(x+a)}}\sim_{a}\frac{1}{\sqrt{2a(x-a)}} $$ and $1/\sqrt{u}$ is convergent as $u\rightarrow 0$. I tried to compute them but failed, what I tried:
- Complex analysis, but because of the sqrt I'm not sure on how to handle it, if ou have any document that explain how to compute this kind of integral, I'll be happy to see it.
- I tried to introduce a new variable $t$ and see the integral as: $$ \int_{a}^{+\infty}\frac{e^{-\sqrt{x^2-a^2}}}{\sqrt{x^2-a^2}}\mathrm{d}x = \int_{a}^{+\infty}\int_{+\infty}^{1}\frac{e^{-t\sqrt{x^2-a^2}}}{\sqrt{x^2-a^2}}\mathrm{d}t\mathrm{d}x$$ and then using substitution $u^2=x^2-a^2$, then I ended up with: $$ \int_{0}^{+\infty}\int_{+\infty}^{1}\frac{e^{-tu}}{\sqrt{u^2+a^2}}\mathrm{d}\mathrm{d}x $$, I gave the $\mathrm{d}t$ integral to sympy that returned me an expression dependant of Meijer function taht I wasn't able to integrate ...
If you have any idea, let me know. Thank you.
The answer is exactly $\color{blue}{\pi H_0^{(1)}(a)}$ for $a>0$, using the Hankel functions. This can be obtained using integral representations of $J_0(a)$ and $Y_0(a)$, with your evaluation continued: \begin{align*} \int_{-\infty}^\infty\frac{e^{i\sqrt{a^2-x^2}}}{\sqrt{a^2-x^2}}\,dx&=2\underbrace{\int_0^a\frac{e^{i\sqrt{a^2-x^2}}}{\sqrt{a^2-x^2}}\,dx}_{x=a\sqrt{1-t^2}}-2i\underbrace{\int_a^\infty\frac{e^{-\sqrt{x^2-a^2}}}{\sqrt{x^2-a^2}}\,dx}_{x=a\sqrt{1+t^2}} \\&=2\int_0^1\frac{e^{iat}\,dt}{\sqrt{1-t^2}}-2i\int_0^\infty\frac{e^{-at}\,dt}{\sqrt{1+t^2}} \\&=\pi\big(J_0(a)+iY_0(a)\big)=\pi H_0^{(1)}(a). \end{align*}