Product of Power Series of Different Powers

I am trying to find the product $M$ of two power series of the form

\begin{equation}
M=\left(\sum_{n=0}^{\infty}a_{n}\, x^{2n}\right)
\left(\sum_{n=0}^{\infty}b_{n}\, x^{n}\right)
\end{equation}

where, $a_{n}=\frac{(-ag^{2})^{n}}{n!}$, and $b_{n}=\frac{(2ag)^{n}}{n!}$.

The product of the two series could be found with the standard formula (discrete convolution) if both series contained powers of $x^{n}$. I have tried to find a way to calculate the product but am not making progress. One potential issues is that $a_{n}$ is alternating and would become imaginary if the square root is taken. How can I calculate this product?

P.S- I suspect the final answer will be an infinite sum over confluent hypergeometric functions.

Additional Information

I am working on an integral of the form

\begin{equation}
\int_{0}^{\infty} x\, e^{-a(gx-b)^{2}}\, e^{-\mu x}\, {_{1}}F_{1}[-\alpha,-\beta,\lambda x] \ dx
\end{equation}

If I keep my limits of integration and write the exponential as a power series I can solve the integral. There is no way I can find to solve the integral if I substitute $u=x-b$. I tried tackling this by writing the exponential in quesiton as:

\begin{equation}
\begin{aligned}
e^{-a(gx-b)^{2}} &= \sum_{n=0}^{\infty}\frac{(-a)^{n}(gx-b)^{2n}}{n!}\\
&= \sum_{n=0}^{\infty}\frac{(-a)^{n}}{n!}\sum_{k=0}^{2n}\binom{2n}{k}(-b)^{2n-k}(gx)^{k}
\end{aligned}
\end{equation}

Switching the order of summation allows for a solution as a single sum:

\begin{equation}
e^{-a(gx-b)^{2}} =\sum_{k=0}^{\infty}\,
\frac{(-a)^{k/2}(-g)^{k}}{\frac{k}{2}!}\,{_{1}}F_{1}\left(\frac{k+1}{2};\frac{1}{2},-ab^{2}\right)\, x^{k}
\end{equation}

This sum has imaginary terms for odd $k$ and is not particularly useful for my purposes.

Answer

We can use the standard formula with a slight variation:

We obtain
\begin{align*}
\left(\sum_{k=0}^\infty a_kx^{2k}\right)\left(\sum_{l=0}^\infty b_lx^l\right)
&=\sum_{n=0}^\infty\left(\sum_{{2k+l=n}\atop{k,l\geq 0}}a_kb_l\right)x^n\tag{1}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}a_kb_{n-2k}\right)x^n\tag{2}
\end{align*}

Comment:

• In (1) the condition for the inner sum is $2k+l=n$ to respect the even powers $x^{2k}$ and all powers $x^l$.




• In (2) we use the floor function to set the upper limit of the inner sum and use $l=n-2k$.