Evaluation of the Integral

Let us consider now the proof that the integral is zero. In order to do this we will separate the circuit in sections and prove the result for each section. The complete circuit $C$ consists of two straight sections $C_{1}$ e $C_{3}$, of the great semicircle $C_{2}$ and of two small semicircles $C_{L}$ e $C_{R}$ of radius $\varepsilon$ around the point $\xi=0$.

For the pair of straight lines $C_{1}$ e $C_{3}$, where we have $d\xi=\imath\, dy$ with $\xi=x+\imath\,y$, taking into account the orientation, we may write

$$I_{C_{1}+C_{3}} = \int_{C_{1}+C_{3}}f(p,\nu,\xi)\,d\xi$$

$$= \imath\, \int_{R}^{\varepsilon}f(p,\nu,\xi)\,dy + \imath\, \int_{-\varepsilon}^{-R}f(p,\nu,\xi)\,dy.$$ (23)

Making in the second integral the transformation of variables $\xi\to-\xi$, which implies $x\to-x$ e $y\to-y$, and since $f(p,\nu ,\xi )$ is odd, we have

$$I_{C_{1}+C_{3}} = 0.$$ (24)

We see therefore that this part of the integral vanishes exactly, independently of the values of $R$ and $\varepsilon$. We are therefore free to take limits involving $R$ or $\varepsilon$ during the calculation of the other sections of the integral, without affecting this result.

Next we consider the two semicircles of radius $\varepsilon$. We will denote this part of the integral, to be calculated according to the criterion of the principal value of Cauchy, as

$$I_{C_{L}+C_{R}} = \int_{C_{L}+C_{R}}f(p,\nu,\xi)\,d\xi.$$ (25)

Since the function $f(p,\nu ,\xi )$ has a simple pole at $\xi=0$, and is also odd, it can be expressed as a Laurent series around this point, with the form

$$f(p,\nu,\xi) = \frac{r_{0}(p,\nu)}{\xi} + \sum_{i=0}^{\infty} c_{i}\xi^{2i+1},$$ (26)

where $r_{0}(p,\nu)$ is the residue of the function at this point, and $c_{i}$ are certain finite coefficients. The series is convergent so long as $\varepsilon$ is smaller than the first zero $\xi_{\nu 1}$. The sum of positive powers represents an analytical function around $\xi=0$, and is therefore regular within the circle of radius $\varepsilon$. It follows that the integral of this regular part goes to zero in the limit $\varepsilon\to 0$, since in this limit both each individual term of the sum and the measure of the domain of integration vanish.

It follows that only the integral of the term containing the pole can remain different from zero in the $\varepsilon\to 0$ limit. We will therefore calculate this integral in polar coordinates, with $\xi=\varepsilon\exp(\imath\,\theta)$ and $d\xi=\imath\,\varepsilon\exp(\imath\,\theta)d\theta$. Since we must use here the Cauchy principal value we have for this part $\Delta I_{C_{L}+C_{R}}$ of the integral $I_{C_{L}+C_{R}}$,

$$\Delta I_{C_{L}+C_{R}} = \frac{1}{2} \int_{C_{L}}\frac{r_{0}(p,\nu)}{\xi}\,d\xi + \frac{1}{2} \int_{C_{R}}\frac{r_{0}(p,\nu)}{\xi}\,d\xi$$

$$= \frac{\imath\,r_{0}(p,\nu)}{2} \int_{\pi/2}^{3\pi/2}\,d\theta + \frac{\imath\,r_{0}(p,\nu)}{2} \int_{\pi/2}^{-\pi/2}\,d\theta$$

$$= 0.$$ (27)

Therefore, this part of the integral $I_{C_{L}+C_{R}}$ also vanishes, and hence the integral $I_{C_{L}+C_{R}}$ vanishes in the limit $\varepsilon\to 0$. Since during this deformation of the circuit no singularities of the function are crossed, and hence the integral does not change, if follows that the integral is zero for all values of $\varepsilon$ smaller than $\xi_{\nu 1}$.

The last section of the circuit we must consider is $C_{2}$. In this case the integral is not zero for finite values of $R$, but we may show that it goes to zero in the limit $R\to\infty$, subject to the condition that for large values of $R$ we have that Eq. (4) holds, so that the circuit does not go over any of the singularities at the points $\xi_{\nu k}$. Using once more polar coordinates, in this section of the circuit we have $d\xi=\imath\,R\exp(\imath\,\theta)d\theta$, where $\xi=R\exp(\imath\,\theta)$, so that the integral is given by

$$I_{C_{2}} = \int_{C_{2}}f(p,\nu,\xi)\,d\xi$$

$$= \imath\,R\int_{-\pi/2}^{\pi/2}d\theta\,e^{\imath\,\theta}\,f(p,\nu,\xi).$$ (28)

Taking the absolute value of the integral and using the triangle inequalities we have

$$\vert I_{C_{2}}\vert \leq R\int_{-\pi/2}^{\pi/2}d\theta\,\vert f(p,\nu,\xi)\vert,$$ (29)

for any value of $R$, and hence also in the $R\to\infty$ limit. We must now consider the behavior of the absolute value of $f(p,\nu ,\xi )$ for large values of $R$. In order to do this we calculate the limit

$$\vert I_{C_{2}}\vert \leq \lim_{R\to\infty} \left[ R\int_{-\pi/2}^{\pi/2}d\theta\, \vert f(p,\nu,\xi)\vert \right]$$

$$= \lim_{R\to\infty} \left[ R\int_{-\pi/2}^{\pi/2}d\theta\, \frac{1}{R^{p+1}}\, \left\vert\frac{J_{\nu+p}(\xi)}{J_{\nu}(\xi)}\right\vert \right].$$ (30)

As we established before, the limit of the absolute value of the ratio of the two Bessel functions is $1$. As a consequence of this, we have for the integral over the section $C_{2}$ of the circuit, in the $R\to\infty$ limit,

$$\vert I_{C_{2}}\vert \leq \lim_{R\to\infty} \left[ \int_{-\pi/2}^{\pi/2}d\theta\, \frac{1}{R^{p}} \right]$$

$$= 0,$$ (31)

since we have $p>0$. This implies, of course, that $I_{C_{2}}=0$ in the $R\to\infty$ limit. We see therefore that the integral of $f(p,\nu ,\xi )$ over the circuit $C$, in the $R\to\infty$ limit, vanishes in all sections of the circuit, and hence that the integral is zero in the $R\to\infty$ limit,

$$\lim_{R\to\infty} \oint_{C}f(p,\nu,\xi)\,d\xi = 0.$$ (32)