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?




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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 $?\|\varphi\|\_{\infty}$.



How to show that this limit converges in the distributional sense to a dirac delta function
Here is an attempt at a solution.


Continuing with your idea: Substitute $\xi:=\frac{b-y}{\sqrt{2}\sigma t^\alpha}$ to make everything nicer:
$$\begin{align}
f\_{t}(\xi) &= \frac{\sqrt{2}C}{\sigma t} \left[\frac{(1-\alpha)}{\alpha} \xi\, \_1F\_1\left(\frac{\alpha+1}{2 \alpha};\frac{3}{2};-\xi^2\right)-\frac{\Gamma \left(\frac{3}{2}-\frac{1}{2 \alpha}\right)}{\Gamma \left(1-\frac{1}{2 \alpha}\right)} {\_1F\_1}\left(\frac{1}{2 \alpha};\frac{1}{2};-\xi^2\right)\right] \tag{1}\\
&= \frac{\sqrt{2}C}{\sigma t} \left[(-\tfrac{1}{2}+\tfrac{1}{2\alpha}) 2\xi {\_1F\_1}\left(\tfrac{1}{2}+\tfrac{1}{2 \alpha};\tfrac{3}{2};-\xi^2\right)-\frac{(\tfrac{1}{2}-\tfrac{1}{2\alpha})\Gamma(\tfrac{1}{2}-\tfrac{1}{2 \alpha})}{\Gamma(1-\frac{1}{2 \alpha})} {\_1F\_1}\left(\tfrac{1}{2 \alpha};\tfrac{1}{2};-\xi^2\right)\right] \tag{2}\\
&= \frac{\sqrt{2}C}{\sigma t} (\tfrac{1}{2}-\tfrac{1}{2\alpha})\left[-2\xi {\_1F\_1}\left(\tfrac{1}{2}+\tfrac{1}{2 \alpha};\tfrac{3}{2};-\xi^2\right)-\frac{\Gamma(\tfrac{1}{2}-\tfrac{1}{2 \alpha})}{\Gamma(1-\frac{1}{2 \alpha})} {\_1F\_1}\left(\tfrac{1}{2 \alpha};\tfrac{1}{2};-\xi^2\right)\right] \tag{3}
\end{align}$$
Now we use the magic formula
$$ {\_1 F\_1}(a;b;-z) = e^{-z} \cdot {\_1 F\_1}(b-a;b;z) \tag{4}$$
and get
$$f\_t(\xi) = \frac{\sqrt{2}C}{\sigma t} (\tfrac{1}{2}-\tfrac{1}{2\alpha})e^{-\xi^2}\left[-2\xi {\_1F\_1}\left(1-\tfrac{1}{2 \alpha};\tfrac{3}{2};\xi^2\right)-\frac{\Gamma(\tfrac{1}{2}-\tfrac{1}{2\alpha})}{\Gamma(1-\frac{1}{2 \alpha})} {\_1F\_1}\left(\tfrac{1}{2}-\tfrac{1}{2\alpha};\tfrac{1}{2};\xi^2\right)\right] \tag{5}$$




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So far not much has happened. But now comes the part where I like the other sign more. If the original function indeed contains a sign error and we are really talking about 
$$f\_t(\xi) = \frac{\sqrt{2}C}{\sigma t} (\tfrac{1}{2}-\tfrac{1}{2\alpha})e^{-\xi^2}\left[-2\xi {\_1F\_1}\left(1-\tfrac{1}{2 \alpha};\tfrac{3}{2};\xi^2\right) \color{red}{+}\frac{\Gamma(\tfrac{1}{2}-\tfrac{1}{2\alpha})}{\Gamma(1-\frac{1}{2 \alpha})} {\_1F\_1}\left(\tfrac{1}{2}-\tfrac{1}{2\alpha};\tfrac{1}{2};\xi^2\right)\right] \tag{6}$$
then we can use generalised Hermite polynomials
$$H\_\nu(z) = 2^\nu \sqrt{\pi} \left(\frac{{\_1 F\_1}(-\frac{\nu}{2};\frac{1}{2};z^2)}{\Gamma(\frac{1-\nu}{2})} - \frac{2z \cdot {\_1 F\_1}(\frac{1-\nu}{2};\frac{3}{2};z^2)}{\Gamma(-\frac{\nu}{2})} \right) \tag{7}$$
with $\nu:=1-\tfrac{1}{\alpha}$ ($\iff \tfrac{1-\nu}{2}=1-\tfrac{1}{2\alpha} \iff -\tfrac{\nu}{2}=\tfrac{1}{2}-\tfrac{1}{2\alpha}$). This gives us:
$$f\_t(\xi) = \frac{\sqrt{2}C}{\sigma t} (-\tfrac{\nu}{2})e^{-\xi^2} \Gamma(-\tfrac{\nu}{2}) 2^{-\nu} \pi^{-1/2} H\_\nu(\xi) \tag{8}$$


In other words: The whole function is a constant multiple of $\tfrac{1}{t}e^{-\xi^2} H\_\nu(\xi)$. Note that this is a true polynomial iff $\alpha\in\{1,\tfrac{1}{2},\tfrac{1}{3},\tfrac{1}{4},\ldots\}$ which may or may not be the case anyway in your situation.


Anyhow: Using the $\xi$-substitution, we find
$$\langle f\_t,\phi\rangle = const \int\_0^\infty \frac{1}{t} e^{-\xi^2} H\_\nu(\xi) \phi(b-\sqrt{2}\sigma t^\alpha \xi) d\xi$$
For $\alpha=1$ this goes to $const\cdot\phi(b)$ so that $f\_t \to const\cdot \delta\_b$ as desired (and the constant will probably work out alright). However, if $\alpha<1$, the integral goes to $sgn(\phi(b))\cdot\infty$.


