Question

Find the asymptotic dependence of the modified Bessel function of the first kind, defined as $$ I_{v}(\alpha) \equiv \frac{1}{2 \pi i} \oint_{C} e^{(\alpha / 2)(z+1 / z)} \frac{d z}{z^{v+1}} $$ where \(C\) starts at \(-\infty\), approaches the origin and circles it, and goes back to \(-\infty\). Thus the negative real axis is excluded from the domain of analyticity.

Short answer

The modified Bessel function of the first kind, \( I_v(\alpha) \), shows an exponential growth for \( |\alpha| \rightarrow \infty \), i.e., has an asymptotic dependence \( e^{|\alpha|} \) for large magnitudes of the argument \( \alpha \).

Step-by-step solution

Integrand Analysis

Rewrite the given integrand \( e^{(\alpha / 2)(z+1 / z)} / z^{v+1} \) as a ratio of two functions \( f(z) = e^{(\alpha / 2)(z+1 / z)} \) and \( g(z) = z^{v+1} \). Then, study the regions where \( f(z) \) and \( g(z) \) are analytic.

Singularities Analysis

Determine the pole of \( 1/z^{v+1} \), which is at \( z=0 \). This pole is avoid by the countour \( C \), so this type of singularity does not contribute to the contour integral. Analysis should reveal that \( f(z) \) contains a branch point at \( z=0 \), which is considered by the contour.

Contour Integral

Substitute \( z=1/w \) into \( I_v(\alpha) \) and differentiate both sides to get two new equivalent expressions for \( I_v(\alpha) \). In each of them, half of the contour integral can be completed by appropriate contour closing at infinity. Evaluate both contour integrals using residue theorem and simplify the sum.

Asymptotic Dependence

Investigate the behaviour of the function for large \( |\alpha| \). Simplify the asymptotic form of \( I_v(\alpha) \) derived in the previous step by factorizing the exponential terms. Take the large \( \alpha \) limit and note that the exponential term \( e^\alpha \) dominates, while all other factors become negligible in such limit.

Key concepts

Contour Integration

Contour integration is a powerful technique in complex analysis, especially for evaluating integrals that are challenging or impossible to compute directly. To understand how to find the asymptotic behaviour of the modified Bessel function of the first kind, we must grasp the basics of contour integration. It involves integrating a complex function around a closed contour or path in the complex plane. The route this path follows, often called contour 'C', can have a significant impact on the value of the integral, depending on the function's behavior along and within the path.

In the example of the modified Bessel function, the path deliberately avoids the negative real axis to exclude any non-analytic behavior from the function being integrated. This technique, paired with an understanding of the function's singularities and analytic regions, allows us to apply powerful theorems such as the residue theorem to simplify and evaluate these complex integrals effectively.

Analytic Functions

Analytic functions, also known as holomorphic functions, are functions that are complex-differentiable in a neighbourhood around every point in their domain. This differentiability implies that these functions are smooth and behave in a predictable manner, which is crucial for contour integration. For a function to be analytic, it must not only be differentiable at a point but must be infinitely differentiable around it and equal to its Taylor series expansion in some neighbourhood.

When we analyze the integrand for the modified Bessel function, we're looking at the properties of two functions, namely, the exponential function and a power function. The exponential function is analytic everywhere except at infinity, while the power function has a pole at zero. The contour 'C' is chosen to navigate these traits, looping around the pole without crossing any non-analytic points, allowing us to take advantage of the integrand's well-behaved nature and evaluate the integral.

Residue Theorem

The residue theorem is a powerful tool in complex analysis that helps compute contour integrals involving analytic functions. It states that if you have a function that is analytic within and on a closed contour 'C', except for a few isolated singularities, the integral of the function around 'C' is equal to 2πi times the sum of the residues at those singularities.

In our case with the modified Bessel function, after manipulating the integral and choosing an appropriate contour, we look for residues, which are essentially coefficients of the function's Laurent series expansion at the singular points. Even though the pole at zero is circumvented by the contour, the method of residues is still applicable to our transformed function. By applying the residue theorem, we're able to convert a complex contour integral into a much simpler calculation, which is a sum involving only the residues at the poles enclosed by the contour. This simplification is a cornerstone in figuring out the behavior of the Bessel function for large values of the parameter α, leading to the desired asymptotic dependence.