Equality of the Schläfli's integrals

Question

I want to show that (ex: (10.9.19) and (10.9.6) in here) $$ J_{\nu}(z) = \frac{1}{2i\pi} \left( \frac{z}{2}\right)^{\nu}\! \int_{\mathcal{H}} e^{w-\frac{z^2}{4\, w}} w^{-\nu-1} \, dw = \frac{1}{\pi} \int_0^\pi \cos(z \sin\theta - \nu \theta)\,d\theta - \frac{\sin \nu \pi}{\pi} \int_0^\infty e^{-z \sinh t - \nu t} \, dt$$ where $\mathcal{H}$ is the Hankel contour (wrapping around the negative half line, because here I take the convention that $w^{\nu-1} :=e^{(\nu -1) \log w}$ with $\ \log (w) = \ln \lvert w\rvert + i \operatorname{Arg}(w),\ \operatorname{Arg}(w) \in ]-\pi,\pi[$, i.e. the principal determination of the logarithm defined on $\mathbb{C}\backslash \mathbb{R}_-$). In the book "Fonctions spéciales de la physique mathématique", A. Lesfari p.79 or in Treatise on the Theory of Bessel Functions, G. N. Watson (Reprint 1996, p.176), the first step is to do the following change of variable $$ w = \frac{z}{2} u\quad \Longrightarrow\quad w -\frac{z^2}{4\, w} = \frac{z}{2} \left( u -\frac{1}{u} \right) \\\text{and if}\; ``w=\gamma(t)= \frac{z}{2} \tilde{\gamma}(t)" \text{then}\ \left(\frac{z}{2}\right)^{\nu} \gamma^{-\nu -1}\, \gamma'\, dt = \left(\frac{z}{2}\right)^{\nu} \frac{d}{d t}\left(\frac{z}{2} \tilde{\gamma}\right)^{-\nu}\! dt = \tilde{\gamma}^{-\nu -1}\, \tilde{\gamma}'\, dt $$ so that $$ J_{\nu}(z) = \frac{1}{2i\pi} \int_{\left(\frac{z}{2}\right)^{-1}\!\cdot\,\mathcal{H}} e^{\frac{z}{2} \left(u-\frac{1}{u} \right)} u^{-\nu-1} \, du$$

It seems to me indeed that $w\in \mathcal{H}\ \Leftrightarrow\ u\in \left(\frac{z}{2}\right)^{-1}\!\cdot\,\mathcal{H}$. They however still integrate over $\mathcal{H}$... what did I missed? It also seems to me that if I integrate over this new contour, I do not get the final result. The problem does not arise in these notes where $z:=x \in \mathbb{R}$.

Answer 1

I'm too lazy to insert the contours only and copy the text, so the beginning will be an extract from some notes I'm writing: the second sentence is a bit awkward but there is no mistake and it was the object of my original question.

What remains to be proved is that $\displaystyle \int_{\mathcal{H}} e^{w-\frac{z^2}{4\, w}}\, \frac{dw}{w^{\nu +1}} = \int_{\tilde{\mathcal{H}}} e^{w-\frac{z^2}{4\, w}}\, \frac{dw}{w^{\nu +1}}$ and similarly for the second deformation (after the rotation=change of variable). Let us start with finite contours, cuting out the part where $\operatorname{Re}(w)< -R$. The difference between the two integrals is an integral over two segments that form the two sides of a triangle which does not contain any pole. By the residue theorem, the integral over the whole triangle vanish, i.e. the difference is the integral along the third side of the triangle, namely $\displaystyle \int_{[-R,-R-iR \tan(\theta_0)]} e^{w-\frac{z^2}{4\, w}}\, \frac{dw}{w^{\nu +1}}$. This is bounded by $$\left\lvert\int_{0_+}^{R\tan (\theta_0)} \frac{e^{(-R-it)-\frac{z^2}{4\hspace{.5pt} (-R-it)}}}{\left(-R-it\right)^{\nu +1}} \, (-i)\, dt \right\rvert \leq \int_{0_+}^{R\tan (\theta_0)} e^{-R + \left\lvert\frac{z}{2}\right\rvert^2 \frac{\cos(2\theta_0) R + \sin (2\theta_0) t}{\left\lvert R + it\right\rvert}} \left\lvert e^{-(\nu + 1)\left(\ln \left\lvert R + it \right\rvert + i \operatorname{Arg}(R+it)\right)} \right\rvert\, dt$$

In the first exponential $\cos(2\theta_0) R + \sin (2\theta_0) t \leq \left\lVert \begin{pmatrix} \cos(2\theta_0)\\ \sin (2\theta_0)\end{pmatrix} \right\rVert_2\, \left\lVert \begin{pmatrix} R \\ t\end{pmatrix}\right\rVert_2 = \left\lvert R+it \right\rvert$ and for the second $\left\lvert e^{\cdots} \right\rvert = \left\lvert R + it \right\rvert^{-\operatorname{Re}(\nu +1)} e^{\operatorname{Im}(\nu + 1)\, \pi} \leq C \left(\frac{R}{\cos \theta_0}\right)^{-\operatorname{Re}(\nu +1)}$ (where I used $\left\lvert R+it \right\rvert \leq \sqrt{R^2 + R^2 \tan^2 \theta_0 } = \frac{R}{\left\lvert \cos \theta_0 \right\rvert}$, $\cos\theta_0$ positive for $\theta_0 < \frac{\pi}{2}$). Finally $$\left\lvert \int_0^{R\tan\theta_0} \cdots dt \right\rvert \leq \int_0^{R\tan (\theta_0)} \tilde{C}\, e^{-R} R^{-\operatorname{Re}(\nu)-1}\, dt = \tilde{C} \tan \theta_0\, e^{-R}\, R^{-\operatorname{Re}(\nu)} \underset{R \to +\infty}{\longrightarrow} 0$$ This means that the difference vanishes when the contours are infinite.

Remark:

1. $\log(w)$ is defined for $w\in \mathbb{C}\backslash \mathbb{R}_-\ \Leftrightarrow\ \tilde{\log}(u)$ is defined for $u= \left(\frac{z}{2}\right)^{-1}\!w \in \mathbb{C}\backslash z^{-1}\mathbb{R}_-$. Those two functions coincide on the sector $\left\lbrace \rho e^{i\theta}\in \mathbb{C},\ -\pi < \theta <\pi-\theta_0 \right\rbrace\ {\color{gray} \ni \left(\frac{z}{2}\right)^{-1}\!\tilde{\mathcal{H}}} $ and is extended back to the principal determination of the $\log$ on $\mathbb{C}\backslash \mathbb{R}_-$.

2. The reason one imposes $\theta_0:=\operatorname{Arg}(z) < \frac{\pi}{2}$ is that if $\tilde{\mathcal{H}}$ when to infinity with infinite $\operatorname{Re}(w)$ then the integral would diverge.

Answer 2

You may approach it in two ways. The first one, you keep a Hankel contour $C´$ with radius 1 in a counterclockwise direction.

Than, you get the following:

For the first piece of contour you get the following integral after applying the substitution $u=\rho e^{-i\pi}$, note that $e^{-i\pi}=-1$;

$$\boxed{L_{1}= \frac{e^{i\pi v}}{2\pi i}\int_{\infty}^{1} e^{-\frac{z}{2}(\rho-\rho^{-1})}\rho^{-(v+1)}d\rho}$$

for the circular contour you make the following substitution $u=e^{i\theta}$: $$L_{2}=\frac{i}{2\pi i}\int_{-\pi}^{\pi} e^{i(z\sin(\theta)-v\theta)}d\theta$$ $$L_{2}=\frac{i}{2\pi i}\int_{-\pi}^{\pi} {\cos(z\sin(\theta)-v\theta)}+i\sin{(z\sin(\theta)-v\theta)}d\theta$$

but since sine in an odd function and the integration interval is symmetric, we get:

$$\boxed{L_{2}=\frac{1}{\pi }\int_{0}^{\pi} {\cos(z\sin(\theta)-v\theta)}d\theta}$$

for the third piece of contour, make the substitution $u=\rho e^{i\pi}$, and you get:

$$\boxed{L_{3}= \frac{e^{-i\pi v}}{2\pi i}\int_{1}^{\infty} e^{-\frac{z}{2}(\rho-\rho^{-1})}\rho^{-(v+1)}d\rho}$$

summing all the pieces together you get: $$\boxed{J_{v}(z)=\frac{1}{\pi }\int_{0}^{\pi} {\cos(z\sin(\theta)-v\theta)}d\theta-\frac{\sin(v \pi)}{\pi}\int_{1}^{\infty} e^{-\frac{z}{2}(\rho-\rho^{-1})}\rho^{-(v+1)}d\rho}$$

Now, let $\rho=e^{w}$ in the last integral and you get your result:

$$\boxed{J_{v}(z)=\frac{1}{\pi }\int_{0}^{\pi} {\cos(z\sin(\theta)-v\theta)}d\theta-\frac{\sin(v \pi)}{\pi}\int_{0}^{\infty} e^{z\sinh(w)}w^{-(v+1)}dw}$$

The second way is similar to the one above, but using the contour below with the appropriate parametrizations, and you get the same result