According to a standard literature
$$\frac{1}{\sqrt{2\pi}q}\int^{-d}_{−∞} \sum ^∞_{k=1} \frac{(A^2/2q^2)^k}{2^k(k!^2)} e^\frac{−r^2}{2q^2}He_{2k}(\frac{r}{q})dr=\frac{1}{\sqrt{2\pi}} \sum ^∞_{k=1} \frac{(A^2/2q^2)^k}{2^k(k!^2)} e^\frac{−d^2}{2q^2}He_{2k}(\frac{d}{q})$$where $$He_n(x)$$ & $$H_n(x)$$is the Hermite Polynomial probabilist's and physicist's convention. My approach is like this:
$$\frac{1}{\sqrt{2\pi}q}\int^{-d}_{−\infty} \sum ^\infty_{k=1} \frac{(A^2/2q^2)^k}{2^k(k!^2)} e^{−r^2/2q^2}He_{2k}(r/q)dr=\frac{1}{\sqrt{2\pi}q}\sum ^\infty_{k=1} \frac{(A^2/2q^2)^k}{2^k(k!^2)}[\int^{0}_{−\infty} e^\frac{−r^2}{2q^2}He_{2k}(\frac{r}{q})dr+\int^{-d}_{0} e^\frac{−r^2}{2q^2}He_{2k}(\frac{r}{q})dr] \tag{1}$$ Converting Hermite Polynomial probabilist's convention to physicist's convention. $$\int^{-d}_{0} e^\frac{−r^2}{2q^2}He_{2k}(\frac{r}{q})dr=\int^{-d}_{0} 2^{-k}e^\frac{−r^2}{2q^2}H_{2k}(\frac{r}{\sqrt{2}q})dr$$
let $$ t=\frac{r}{\sqrt{2}q} $$ then changing integration variable and limits we get $$\int^{-d}_{0} 2^{-k}e^\frac{−r^2}{2q^2}H_{2k}(\frac{r}{\sqrt{2}q})dr=2^{-k}\sqrt{2}q\int^{\frac{-d}{\sqrt{2}q}}_{0} e^{-t^2}H_{2k}(t)dt=2^{-k}\sqrt{2}q[H_{2k-1}(0)- e^{-d^2/2q^2}H_{2k-1}(-d/\sqrt{2}q)]=2^{-k}\sqrt{2}q e^{-d^2/2q^2}H_{2k-1}(d/\sqrt{2}q) $$
$$H_{2n-1}(0)=0 \qquad\text{and}\qquad H_{n}(-x)=(-1)^nH_{n}(x)$$
Going back to the probabilist's form $$2^{-k}\sqrt{2}q e^{-d^2/2q^2}H_{2k-1}(d/\sqrt{2}q)=q e^{-d^2/2q^2}He_{2k-1}(d/q)$$ Then from (1) $$\frac{1}{\sqrt{2\pi}q}\sum ^∞_{k=1} \frac{(A^2/2q^2)^k}{2^k(k!^2)}\int^{-d}_{0} e^{−r^2/2q^2}He_{2k}(r/q)dr=\frac{1}{\sqrt{2\pi}q}\sum ^∞_{k=1} \frac{(A^2/2q^2)^k}{2^k(k!^2)}q e^{-d^2/2q^2}He_{2k-1}(d/q)=\frac{1}{\sqrt{2\pi}}\sum ^∞_{k=1} \frac{(A^2/2q^2)^k}{2^k(k!^2)} e^{-d^2/2q^2}He_{2k-1}(d/q)$$
I have carried out calculation with the second integral of rhs of (1) and obtained the result. My question is whether the first integral of rhs produces a zero value? Please comment if any mistake is found.