Let $$\begin{eqnarray}\nonumber f(y, t) &=& \frac{C}{\sigma ^2 t} \left[\frac{(1-\alpha) (b-y)}{\alpha t^{\alpha}} \, _1F_1\left[\frac{\alpha+1}{2 \alpha};\frac{3}{2};-\frac{ (b-y)^2}{2 \sigma^2 t^{2 \alpha}}\right]- \sqrt{2} \sigma \frac{\Gamma \left(\frac{3}{2}-\frac{1}{2 \alpha}\right)}{\Gamma \left(1-\frac{1}{2 \alpha}\right)} \, _1F_1\left[\frac{1}{2 \alpha};\frac{1}{2};-\frac{ (b-y)^2}{2 \sigma^2 t^{2 \alpha}}\right]\right] \end{eqnarray}$$ be defined in $f(y, t) \in (-\infty, b]$, where $0 < \alpha < 1$, $\sigma > 0$ and $t$ represents time and $C$ is some scalar value. Also where $_1F_1(a,b,z)$ denotes the Kummer confluent hypergeometric function with series expansion $$_1F_1 (a,b,z)=\sum _{k=0}^{\infty } \frac{(a)_k}{(b)_k}\left(\frac{z^k}{k!}\right),$$ where $(c)_k$ denote Pochhammer’s Symbol \begin{eqnarray}\nonumber (c)_0 &=& 1\\\nonumber (c)_n &=&c(c+1)(c+2)⋯(c + n -1),\\\nonumber (c)_n &=&\frac{\Gamma(c + n)}{\Gamma(c)}\quad c \neq 0, -1, -2, \cdots \end{eqnarray} and $\Gamma(z)$ denotes the Gamma function which satisfies $$\Gamma (z)=\int _0^{\infty } e^{-t} t^{z-1} dt.$$

I am interested in showing convergence in the distributional sense, that for some $y_0$ the following limit holds $$\lim_{t \rightarrow 0}f\left(y, t\right) = \delta\left(y - y_0\right), \quad y_0 \in \left(- \infty, b\right).$$

The presence of Kummer confluent hypergeometric function have made evaluating Fourier transform of $f(y,t)$ or applying dominated convergence theorem too hard. I am having trouble proving this analytically. Any help?

Some observations which may be helpful $$\int_{-\infty}^b f(y, t) \mathrm{d}y = C t^{\alpha-1}.$$ $$\int_{-\infty}^{\infty} f(y, t) \mathrm{d}y = 0.$$ Contour integral representation of Kummer confluent hypergeometric function $$_1F_1 (a,b,z)= \frac{\Gamma (b)}{2 \pi \iota \Gamma (a)} \int_{\gamma -\iota\infty}^{\gamma +\iota\infty} \frac{(-z)^{-s} \Gamma (s) \Gamma (a-s)}{\Gamma (b-s)}\mathrm{d}s$$ where $ 0 <\gamma < Re(a) \wedge |arg(-z)| < \frac{\pi}{2}.$

Denoting $$ I(a,b,z) = \int_{\gamma -\iota\infty}^{\gamma +\iota\infty} \frac{(-z)^{-s} \Gamma (s) \Gamma (a-s)}{\Gamma (b-s)}\mathrm{d}s$$ Then the function of interest can be rewritten as

## \begin{eqnarray} f(y, t) &=& \frac{C}{\sigma ^2 t} [\frac{\sqrt{\pi }(1-\alpha) (b-y)}{2 \alpha t^{\alpha} \Gamma \left(\frac{\alpha+1}{2 \alpha}\right)} \, I\left[\frac{\alpha+1}{2 \alpha};\frac{3}{2};-\frac{ (b-y)^2}{2 \sigma^2 t^{2 \alpha}}\right] \\&-& \sqrt{\frac{2}{\pi}} \sigma \sin{\left(\frac{\pi }{2 H}\right)} \Gamma \left(\frac{3}{2}-\frac{1}{2 \alpha}\right) I\left[\frac{1}{2 \alpha};\frac{1}{2};-\frac{ (b-y)^2}{2 \sigma^2 t^{2 \alpha}}\right]] \end{eqnarray}

**Solution attempt:**

Let $\varphi \in C_{c}^{\infty}(\mathbb{R})$. Using the substitution $x = \frac{b-y}{\sigma t^\alpha}$ we obtain \begin{eqnarray}\nonumber f_{t}(x) &=& \frac{C}{\sigma t} \left[\frac{(1-\alpha)x}{\alpha} \, _1F_1\left[\frac{\alpha+1}{2 \alpha};\frac{3}{2};-\frac{x^2}{2}\right]- \sqrt{2} \frac{\Gamma \left(\frac{3}{2}-\frac{1}{2 \alpha}\right)}{\Gamma \left(1-\frac{1}{2 \alpha}\right)} \, _1F_1\left[\frac{1}{2 \alpha};\frac{1}{2};-\frac{ x^2}{2}\right]\right]\\ \langle f_{t},\varphi \rangle &=& \int_{-\infty}^{\infty} I_{\left(-\infty, b\right]} f(y,t) \varphi(y) \mathrm{d}y \\ &=& -\sigma t^\alpha\int_{-\infty}^{\infty} I_{\left[0, \infty\right)} f_{t}(x) \varphi(b - x \sigma t^{\alpha}) \mathrm{d}x \end{eqnarray} As $\varphi$ is continuous and compactly supported, this integrand is dominated by $? \|\varphi\|_{\infty}$ which integrates to $?\|\varphi\|_{\infty}$.

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