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A.2. The confluent hypergeometric functions [056T]

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A.2. The confluent hypergeometric functions

Now we summarize some results regarding the confluent hypergeometric functions which are used in Section 5. Given α,β∈ℝ\alpha,\beta\in\mathbb{R} such that α>β\alpha>\beta and α\alpha is not a negative integer, we consider the following confluent hypergeometric equation

(A.14) y⋅d2​𝒥​(y)d​y2+(α−y)⋅d​𝒥​(y)dy−β⋅𝒥⁡(y)=0.y\cdot\frac{d^{2}\mathcal{J}(y)}{dy^{2}}+(\fa-y)\cdot\frac{d\mathcal{J}(y)}{dy}-\fb\cdot\mathcal{J}(y)=0.

Let

(A.15) Φ♯⁡(β,α,y)≡∑k=0∞(β)k(α)k⋅ykk!,\Ku(\beta,\alpha,y)\equiv\sum\limits_{k=0}^{\infty}\frac{(\beta)_{k}}{(\alpha)_{k}}\cdot\frac{y^{k}}{k!},

where we define the notation (x)k≡∏m=1k(x+m−1)(x)_{k}\equiv\prod\limits_{m=1}^{k}(x+m-1) and (x)0=1(x)_{0}=1. So the power series Φ♯⁡(β,α,z)\Ku(\fb,\fa,z) is always well-defined for all β∈ℂ\fb\in\mathbb{C}, z∈ℂz\in\mathbb{C} and α∈ℂ∖{0,−1,−2,…}\fa\in\mathbb{C}\setminus\{0,-1,-2,\ldots\}. Moreover, for any fixed z∈ℂz\in\mathbb{C}, the function Φ♯\Ku is entire in β\fb and meromorphic in α\fa with simple poles at negative integers.

It is by straightforward calculations that the function Φ♯⁡(β,α,y)\Ku(\fb,\fa,y) is a solution to (A.14). In the literature, Φ♯\Ku is called Kummer’s (confluent hypergeometric) function. Moreover, when y>0y>0, one can directly check that the function Φ♯^​(β,α,y)≡y1−α⋅Φ♯⁡(1+β−α,2−α,y)\widehat{\Ku}(\fb,\fa,y)\equiv y^{1-\fa}\cdot\Ku(1+\fb-\fa,2-\fa,y), which is linearly independent of Φ♯⁡(β,α,y)\Ku(\fb,\fa,y), also solves (A.14). Therefore, the general solution of (A.14) for y>0y>0 is

(A.16) 𝒥⁡(y)=C⋅Φ♯⁡(β,α,y)+C∗⋅y1−α⋅Φ♯⁡(1+β−α,2−α,y).\mathcal{J}(y)=C\cdot\Ku(\fb,\fa,y)+C^{*}\cdot y^{1-\fa}\cdot\Ku(1+\fb-\fa,2-\fa,y).

The power series definition of Φ♯⁡(β,α,y)\Ku(\fb,\fa,y) immediately gives the following integral representation formula which is well known in the literature. We include a short proof just for the convenience of the readers.

Lemma A.2.

For any α>β>0\fa>\fb>0, then for each y∈ℝy\in\mathbb{R},

(A.17) Φ♯⁡(β,α,y)=Γ⁡(α)Γ⁡(β)​Γ​(α−β)​∫01eyt​tβ−1​(1−t)α−β−1​dt.\Ku(\fb,\fa,y)=\frac{\Gamma(\fa)}{\Gamma(\fb)\Gamma(\fa-\fb)}\int_{0}^{1}e^{yt}t^{\fb-1}(1-t)^{\fa-\fb-1}dt.
Proof.

Given p,q>0p,q>0, let B⁡(p,q)B(p,q) be the beta function which is defined by

(A.18) B⁡(p,q)≡∫01tp−1​(1−t)q−1​𝑑t.B(p,q)\equiv\int_{0}^{1}t^{p-1}(1-t)^{q-1}dt.

Then the beta function satisfies B⁡(p,q)=Γ⁡(p)​Γ​(q)Γ⁡(p+q)B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}. The above formulae imply that

(A.19) (β)k(α)k\displaystyle\frac{(\fb)_{k}}{(\fa)_{k}} =\displaystyle= Γ⁡(β+k)Γ⁡(β)⋅Γ⁡(α)Γ⁡(α+k)\displaystyle\frac{\Gamma(\fb+k)}{\Gamma(\fb)}\cdot\frac{\Gamma(\fa)}{\Gamma(\fa+k)}
=\displaystyle= Γ⁡(α)Γ⁡(β)⋅B⁡(β+k,α−β)Γ⁡(α−β)\displaystyle\frac{\Gamma(\fa)}{\Gamma(\fb)}\cdot\frac{B(\fb+k,\fa-\fb)}{\Gamma(\fa-\fb)}
=\displaystyle= Γ⁡(α)Γ⁡(β)​Γ​(α−β)​∫01tβ+k−1​(1−t)α−β−1​𝑑t.\displaystyle\frac{\Gamma(\fa)}{\Gamma(\fb)\Gamma(\fa-\fb)}\int_{0}^{1}t^{\fb+k-1}(1-t)^{\fa-\fb-1}dt.

Now we return to the definition of Φ♯\Ku, combining the above summation,

(A.20) Φ♯⁡(β,α,y)\displaystyle\Ku(\beta,\alpha,y) =\displaystyle= ∑k=0∞(β)k(α)k⋅ykk!\displaystyle\sum\limits_{k=0}^{\infty}\frac{(\beta)_{k}}{(\alpha)_{k}}\cdot\frac{y^{k}}{k!}
=\displaystyle= Γ⁡(α)Γ⁡(β)​Γ​(α−β)​∫01tβ−1​(1−t)α−β−1​∑k=0∞(y​t)k−1k!​𝑑t\displaystyle\frac{\Gamma(\fa)}{\Gamma(\fb)\Gamma(\fa-\fb)}\int_{0}^{1}t^{\beta-1}(1-t)^{\fa-\fb-1}\sum\limits_{k=0}^{\infty}\frac{(yt)^{k-1}}{k!}dt
=\displaystyle= Γ⁡(α)Γ⁡(β)​Γ​(α−β)​∫01ey​t​tβ−1​(1−t)α−β−1​𝑑t.\displaystyle\frac{\Gamma(\fa)}{\Gamma(\fb)\Gamma(\fa-\fb)}\int_{0}^{1}e^{yt}t^{\beta-1}(1-t)^{\fa-\fb-1}dt.

The proof is done.

∎

Given β>0\fb>0 and y>0y>0, we define the function

(A.21) 𝒰⁡(β,α,y)≡1Γ⁡(β)​∫0∞e−yt​tβ−1​(1+t)α−β−1​dt.\mathcal{U}(\fb,\fa,y)\equiv\frac{1}{\Gamma(\fb)}\int_{0}^{\infty}e^{-yt}t^{\fb-1}(1+t)^{\fa-\fb-1}dt.

Quick computations show that for each β>0\fb>0, the function 𝒰⁡(β,α,y)\mathcal{U}(\fb,\fa,y) is a solution to the confluent hypergeometric equation (A.14) on the positive real axis ℝ+\mathbb{R}_{+}. Now let β>0\fb>0 and α∈ℝ∖{0,−1,−2,−3,…}\fa\in\mathbb{R}\setminus\{0,-1,-2,-3,\ldots\}, thanks to (A.16), the function 𝒰⁡(β,α,y)\mathcal{U}(\fb,\fa,y) can be written in terms of Kummer’s function Φ♯\Ku. Evaluating those functions and their derivatives at y=0y=0, one can easily obtain

(A.22) 𝒰⁡(β,α,y)=Γ⁡(1−α)Γ⁡(1+β−α)⋅Φ♯⁡(β,α,y)+Γ⁡(α−1)Γ⁡(β)⋅y1−α⋅Φ♯⁡(1+β−α,2−α,y).\mathcal{U}(\fb,\fa,y)=\frac{\Gamma(1-\fa)}{\Gamma(1+\fb-\fa)}\cdot\Ku(\fb,\fa,y)+\frac{\Gamma(\fa-1)}{\Gamma(\fb)}\cdot y^{1-\fa}\cdot\Ku(1+\fb-\fa,2-\fa,y).

Notice that, the above relation is well-defined for each y≥0y\geq 0 and non-integral α\alpha. Moreover, if α→n+1∈ℤ+\alpha\to n+1\in\mathbb{Z}_{+}, then the right hand side of (A.22) will tend to a definite limit. The function 𝒰⁡(β,α,y)\mathcal{U}(\fb,\fa,y) is usually called Tricomi’s (confluent hypergeometric) function. In our context, we are also interested in the case y<0y<0. It can be directly verified that, if y<0y<0, the function

(A.23) Ψ♭⁡(β,α,y)≡ey⋅𝒰⁡(α−β,α,−y)\Tri(\fb,\fa,y)\equiv e^{y}\cdot\mathcal{U}(\fa-\fb,\fa,-y)

solves equation (A.14). Moreover, it immediately follows from the integral representation of 𝒰\mathcal{U} that for any y<0y<0,

(A.24) Ψ♭⁡(β,α,y)=eyΓ⁡(α−β)​∫0∞eyt​tα−β−1​(1+t)β−1​dt.\Tri(\beta,\alpha,y)=\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{\infty}e^{yt}t^{\fa-\fb-1}(1+t)^{\fb-1}dt.

In summary, if y<0y<0, the equation (A.14) has two linearly independent solutions Φ♯⁡(β,α,y)\Ku(\fb,\fa,y) and Ψ♭⁡(β,α,y)\Tri(\fb,\fa,y).

The asymptotic behavior of Φ♯⁡(β,α,y)\Ku(\fb,\fa,y), 𝒰⁡(β,α,y)\mathcal{U}(\fb,\fa,y) and Ψ♭⁡(β,α,y)\Tri(\fb,\fa,y) can be easily seen from the above integral formulae. In fact, we have the following

Lemma A.3.

The following asymptotics hold:

  1. (1)

    Let α∈ℝ∖{0,−1,−2,−3,…}\fa\in\mathbb{R}\setminus\{0,-1,-2,-3,\ldots\} and β>0\fb>0 satisfy α>β+1\fa>\fb+1, then

    (A.25) Φ♯⁡(β,α,y)∼{Γ⁡(α)Γ⁡(α−β)⋅(−y)−β,y→−∞,Γ⁡(α)Γ⁡(β)⋅ey⋅yβ−α,y→+∞.\displaystyle\Ku(\fb,\fa,y)\sim\begin{cases}\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}\cdot(-y)^{-\beta},&y\to-\infty,\\ \frac{\Gamma(\fa)}{\Gamma(\fb)}\cdot e^{y}\cdot y^{\fb-\fa},&y\to+\infty.\end{cases}
  2. (2)

    Let β>0\beta>0, then

    (A.26) 𝒰(β,α,y)∼y−β,y→+∞.\mathcal{U}(\fb,\fa,y)\sim y^{-\fb},\ y\to+\infty.
  3. (3)

    Let α>β\alpha>\beta, then

    (A.27) Ψ♭⁡(β,α,y)∼ey⋅(−y)β−α,y→−∞.\Tri(\fb,\fa,y)\sim e^{y}\cdot(-y)^{\fb-\fa},\ y\to-\infty.
Proof.

The proof is straightforward. For example, we only prove

(A.28) Φ♯⁡(β,α,y)∼Γ⁡(α)Γ⁡(α−β)⋅(−y)−β\Ku(\fb,\fa,y)\sim\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}\cdot(-y)^{-\beta}

as y→−∞y\to-\infty. The calculations of the remaining cases are the same. We make change of variables and let u=−y​tu=-yt, then

Φ♯⁡(β,α,y)\displaystyle\Ku(\fb,\fa,y) =Γ⁡(α)Γ⁡(β)​Γ​(α−β)​∫01ey​t​tβ−1​(1−t)α−β−1​𝑑t\displaystyle=\frac{\Gamma(\fa)}{\Gamma(\fb)\Gamma(\fa-\fb)}\int_{0}^{1}e^{yt}t^{\fb-1}(1-t)^{\fa-\fb-1}dt
(A.29) =Γ⁡(α)Γ⁡(β)​Γ​(α−β)⋅(−y)−β⋅∫0−ye−uuβ−1(1+uy)α−β−1du.\displaystyle=\frac{\Gamma(\fa)}{\Gamma(\fb)\Gamma(\fa-\fb)}\cdot(-y)^{-\fb}\cdot\int_{0}^{-y}e^{-u}u^{\fb-1}\Big(1+\frac{u}{y}\Big)^{\fa-\fb-1}du.

Since α−β−1>0\fa-\fb-1>0 and −1≤uy≤0-1\leq\frac{u}{y}\leq 0, it is obvious (1+uy)α−β−1≤1(1+\frac{u}{y})^{\fa-\fb-1}\leq 1. Hence dominated convergence theorem implies

(A.30) limy→−∞∫0−ye−u​uβ−1​(1+uy)α−β−1​𝑑u=∫0∞e−u​uβ−1​𝑑u=Γ⁡(β).\lim\limits_{y\to-\infty}\int_{0}^{-y}e^{-u}u^{\fb-1}\Big(1+\frac{u}{y}\Big)^{\fa-\fb-1}du=\int_{0}^{\infty}e^{-u}u^{\fb-1}du=\Gamma(\beta).

Therefore, as y→−∞y\to-\infty,

(A.31) Φ♯(β,α,y)∼Γ⁡(α)Γ⁡(α−β)⋅(−y)−β.\Ku(\fb,\fa,y)\sim\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot(-y)^{-\fb}.

∎

Next we introduce some recurrence formulae for Kummer’s function.

Lemma A.4.

Let α∈ℝ∖{0,−1,−2,−3,…}\fa\in\mathbb{R}\setminus\{0,-1,-2,-3,\ldots\} and β∈ℝ\fb\in\mathbb{R}, then for each y∈ℝy\in\mathbb{R},

(A.32) Φ♯⁡(β,α,y)\displaystyle\Ku(\fb,\fa,y) =Φ♯⁡(β+1,α,y)−yα​Φ♯⁡(β+1,α+1,y),\displaystyle=\Ku(\fb+1,\fa,y)-\frac{y}{\fa}\Ku(\fb+1,\fa+1,y),
(A.33) Φ♯⁡(β,α,y)\displaystyle\Ku(\fb,\fa,y) =α+yα⋅Φ♯⁡(β,α+1,y)−α−β+1α⁡(α+1)⋅y⋅Φ♯⁡(β,α+1,y).\displaystyle=\frac{\fa+y}{\fa}\cdot\Ku(\fb,\fa+1,y)-\frac{\fa-\fb+1}{\fa(\fa+1)}\cdot y\cdot\Ku(\fb,\fa+1,y).
Proof.

The formula can be quickly verified by applying the power series definition of Φ♯\Ku. ∎

With the above recurrence formula, we can extend the domain of indices in Lemma A.3 for Kummer’s function.

Lemma A.5.

For any α∈ℝ∖{0,−1,−2,−3,…}\fa\in\mathbb{R}\setminus\{0,-1,-2,-3,\ldots\} and β∈ℝ\beta\in\mathbb{R} such that α>β\fa>\fb, then

(A.34) Φ♯⁡(β,α,y)∼{Γ⁡(α)Γ⁡(α−β)⋅(−y)−β,y→−∞,Γ⁡(α)Γ⁡(β)⋅ey⋅yβ−α,y→+∞.\displaystyle\Ku(\fb,\fa,y)\sim\begin{cases}\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}\cdot(-y)^{-\beta},&y\to-\infty,\\ \frac{\Gamma(\fa)}{\Gamma(\fb)}\cdot e^{y}\cdot y^{\fb-\fa},&y\to+\infty.\end{cases}
Proof.

We start with the initial step by assuming α−β>1\alpha-\beta>1 and β>1\beta>1. Then Lemma A.3 in this case shows that the desired asymptotics hold in this case.

Applying the recurrence formula (A.33), we can extend the domain of indices to α−β>0\alpha-\beta>0 and β>1\beta>1. Then applying (A.32), one can obtain the desired asymptotics for all β∈ℝ\beta\in\mathbb{R}. The proof is done. ∎

Lemma A.6 (Kummer’s transformation law).

Let α∈ℝ∖{0,−1,−2,−3,…}\fa\in\mathbb{R}\setminus\{0,-1,-2,-3,\ldots\} and β∈ℝ\fb\in\mathbb{R}, then for any y∈ℝy\in\mathbb{R},

(A.35) Φ♯⁡(β,α,y)=ey⋅Φ♯⁡(α−β,α,−y).\Ku(\fb,\fa,y)=e^{y}\cdot\Ku(\fa-\fb,\fa,-y).
Proof.

First, we temporarily assume α>β>0\fa>\fb>0. By Lemma A.2,

(A.36) ey⋅Φ♯⁡(α−β,α,−y)\displaystyle e^{y}\cdot\Ku(\fa-\fb,\fa,-y) =\displaystyle= Γ⁡(α)Γ⁡(α−β)​Γ​(β)​∫01ey⁡(1−t)​tα−β−1​(1−t)β−1​𝑑t\displaystyle\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)\Gamma(\fb)}\int_{0}^{1}e^{y(1-t)}t^{\fa-\fb-1}(1-t)^{\fb-1}dt
=\displaystyle= Γ⁡(α)Γ⁡(α−β)​Γ​(β)​∫01ey​s​(1−s)α−β−1​sβ−1​𝑑s\displaystyle\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)\Gamma(\fb)}\int_{0}^{1}e^{ys}(1-s)^{\fa-\fb-1}s^{\fb-1}ds
=\displaystyle= Φ♯⁡(β,α,y).\displaystyle\Ku(\fb,\fa,y).

Now we prove the general case. Since both ey⋅Φ♯⁡(α−β,α,−y)Γ⁡(α)\frac{e^{y}\cdot\Ku(\fa-\fb,\fa,-y)}{\Gamma(\fa)} and Φ♯⁡(β,α,y)Γ⁡(α)\frac{\Ku(\fb,\fa,y)}{\Gamma(\fa)} are entire functions in ℂ\mathbb{C}, so the standard analytic continuation theorem implies that Φ♯⁡(β,α,y)=ey⋅Φ♯⁡(α−β,α,−y)\Ku(\fb,\fa,y)=e^{y}\cdot\Ku(\fa-\fb,\fa,-y) holds for any arbitrary β∈ℝ\beta\in\mathbb{R} and α∈ℝ∖{0,−1,−2,−3,…}\alpha\in\mathbb{R}\setminus\{0,-1,-2,-3,\ldots\}. ∎

Next we give another integral representation for Kummer’s function Φ♯⁡(β,α,y)\Ku(\fb,\fa,y) in the case y≤0y\leq 0, which has a crucial role in Section 5.

Lemma A.7.

Assume that α>β\fa>\fb and y≤0y\leq 0, then it holds that

(A.37) Φ♯⁡(β,α,y)=Γ⁡(α)Γ⁡(α−β)⋅ey​(−y)1−α2⋅∫0∞e−t⋅tα−12−β⋅Iα−1​(2​−yt)​dt.\Ku(\fb,\fa,y)=\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot e^{y}(-y)^{\frac{1-\fa}{2}}\cdot\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt.
Proof.

By definition,

(A.38) Iα−1​(2​−y​t)=∑k=0∞(−y​t)k+α−12k!⋅Γ⁡(k+α).I_{\fa-1}(2\sqrt{-yt})=\sum\limits_{k=0}^{\infty}\frac{(-yt)^{k+\frac{\fa-1}{2}}}{k!\cdot\Gamma(k+\fa)}.

Integrating the above expansion, it follows that

(A.39) Γ⁡(α)Γ⁡(α−β)⋅∫0∞e−t⋅tα−12−β⋅Iα−1​(2​−y​t)​𝑑t\displaystyle\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt
=\displaystyle= Γ⁡(α)Γ⁡(α−β)⋅(−y)α−12⋅∑k=0∞(−y)kk!⋅Γ⁡(k+α)⋅∫0∞e−t⋅tα−β+k−1​𝑑t\displaystyle\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot(-y)^{\frac{\alpha-1}{2}}\cdot\sum\limits_{k=0}^{\infty}\frac{(-y)^{k}}{k!\cdot\Gamma(k+\fa)}\cdot\int_{0}^{\infty}e^{-t}\cdot t^{\fa-\fb+k-1}dt
=\displaystyle= Γ⁡(α)Γ⁡(α−β)⋅(−y)α−12⋅∑k=0∞(−y)k⋅Γ⁡(α−β+k)k!⋅Γ⁡(k+α).\displaystyle\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot(-y)^{\frac{\alpha-1}{2}}\cdot\sum\limits_{k=0}^{\infty}\frac{(-y)^{k}\cdot\Gamma(\alpha-\beta+k)}{k!\cdot\Gamma(k+\fa)}.

By the recursive formula of the Gamma function, Γ⁡(α−β+k)Γ⁡(k+α)=(α−β)k⋅Γ⁡(α−β)(α)k⋅Γ⁡(α)\frac{\Gamma(\alpha-\beta+k)}{\Gamma(k+\alpha)}=\frac{(\alpha-\beta)_{k}\cdot\Gamma(\alpha-\beta)}{(\alpha)_{k}\cdot\Gamma(\alpha)}, so it follows that

(A.40) Γ⁡(α)Γ⁡(α−β)⋅∫0∞e−t⋅tα−12−β⋅Iα−1​(2​−y​t)​𝑑t\displaystyle\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt
=\displaystyle= (−y)α−12⋅∑k=0∞(α−β)k​(−y)k(α)k⋅k!\displaystyle(-y)^{\frac{\alpha-1}{2}}\cdot\sum\limits_{k=0}^{\infty}\frac{(\alpha-\beta)_{k}(-y)^{k}}{(\alpha)_{k}\cdot k!}
=\displaystyle= (−y)α−12⋅Φ♯⁡(α−β,α,−y).\displaystyle(-y)^{\frac{\alpha-1}{2}}\cdot\Ku(\fa-\fb,\fa,-y).

Therefore,

(A.41) Γ⁡(α)Γ⁡(α−β)⋅ey​(−y)1−α2⋅∫0∞e−t⋅tα−12−β⋅Iα−1​(2​−y​t)​𝑑t\displaystyle\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot e^{y}(-y)^{\frac{1-\fa}{2}}\cdot\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt
=\displaystyle= ey⋅Φ♯⁡(α−β,α,−y)\displaystyle e^{y}\cdot\Ku(\fa-\fb,\fa,-y)
=\displaystyle= Φ♯⁡(β,α,y).\displaystyle\Ku(\fb,\fa,y).

The last equality follows from Kummer’s transformation law.

∎

Lemma A.8.

Let ν>0\nu>0, then for all y>0y>0

(A.42) Iν​(y)\displaystyle I_{\nu}(y) =(y2)ν​e−yΓ⁡(ν+1)​Φ♯⁡(ν+12,2​ν+1,2​y),\displaystyle=\frac{(\frac{y}{2})^{\nu}e^{-y}}{\Gamma(\nu+1)}\Ku(\nu+\frac{1}{2},2\nu+1,2y),
(A.43) Kν​(y)\displaystyle K_{\nu}(y) =π​(2​y)ν​e−y​𝒰​(ν+12,2​ν+1,2​y).\displaystyle=\sqrt{\pi}(2y)^{\nu}e^{-y}\mathcal{U}(\nu+\frac{1}{2},2\nu+1,2y).
Proof.

The relation (A.42) can be verified by the power series definition of IνI_{\nu} and Φ♯⁡(ν+12,2​ν+1,2​y)\Ku(\nu+\frac{1}{2},2\nu+1,2y), so we just omit the computations.

To prove (A.43), first we assume ν\nu is not an integer. Combining the definition

(A.44) Kν​(y)=πsin⁡(ν​π)⋅I−ν​(y)−Iν​(y)2K_{\nu}(y)=\frac{\pi}{\sin(\nu\pi)}\cdot\frac{I_{-\nu}(y)-I_{\nu}(y)}{2}

and the relation

(A.45) 𝒰⁡(ν+12,2​ν+1,y)=Γ⁡(−2​ν)Γ⁡(12−ν)⋅Φ♯⁡(ν+12,2​ν+1,y)+Γ⁡(2​ν)Γ⁡(ν+12)⋅y−2​ν⋅Φ♯⁡(12−ν,1−2​ν,y),\mathcal{U}(\nu+\frac{1}{2},2\nu+1,y)=\frac{\Gamma(-2\nu)}{\Gamma(\frac{1}{2}-\nu)}\cdot\Ku(\nu+\frac{1}{2},2\nu+1,y)+\frac{\Gamma(2\nu)}{\Gamma(\nu+\frac{1}{2})}\cdot y^{-2\nu}\cdot\Ku(\frac{1}{2}-\nu,1-2\nu,y),

which is given by (A.22). If ν\nu is an integer, the relation (A.43) can be obtained by the limiting definition of KνK_{\nu} and the continuity argument for ν\nu.

∎

The following corollary shows the asymptotic behavior of Iν​(y)I_{\nu}(y) and Kν​(y)K_{\nu}(y) as y→+∞y\to+\infty.

Corollary A.8.1.

Let ν>0\nu>0, then we have

(A.46) limy→+∞Iν​(y)ey2​π​y=1\lim\limits_{y\to+\infty}\frac{I_{\nu}(y)}{\frac{e^{y}}{\sqrt{2\pi y}}}=1

and

(A.47) limy→+∞Kν​(y)π2​y⋅e−y=1.\lim\limits_{y\to+\infty}\frac{K_{\nu}(y)}{\sqrt{\frac{\pi}{2y}}\cdot e^{-y}}=1.
Proof.

The proof follows from Lemma A.3, Lemma A.5 and Lemma A.8. ∎

Original mathematics by the credited authors. Source-backed reader collection; mathematical self-containment is not assessed.