1. **State the problem:** Evaluate the integral $$\frac{1}{\sqrt{2\pi}}\int \left( \frac{d^m}{dq^m} e^{-q^2} \right) e^{q^2} e^{-i s q} \, dq$$ 2. **Rewrite the integrand:** Note that inside the integral we have the $m$-th derivative of $e^{-q^2}$ multiplied by $e^{q^2}$, which simplifies the expression inside the integral: $$\left( \frac{d^m}{dq^m} e^{-q^2} \right) e^{q^2} = \frac{d^m}{dq^m} e^{-q^2} \cdot e^{q^2}$$ 3. **Use the Leibniz rule and properties of derivatives:** Since $e^{-q^2} e^{q^2} = 1$, the expression simplifies to the $m$-th derivative of $1$ times $e^{-i s q}$, but we must be careful because the derivative acts only on $e^{-q^2}$. 4. **Recognize the Hermite polynomial relation:** Recall the physicist's Hermite polynomials $H_m(q)$ satisfy: $$H_m(q) = (-1)^m e^{q^2} \frac{d^m}{dq^m} e^{-q^2}$$ Rearranged: $$\frac{d^m}{dq^m} e^{-q^2} = (-1)^m e^{-q^2} H_m(q)$$ 5. **Substitute back into the integral:** $$\frac{1}{\sqrt{2\pi}} \int \left( \frac{d^m}{dq^m} e^{-q^2} \right) e^{q^2} e^{-i s q} \, dq = \frac{1}{\sqrt{2\pi}} \int (-1)^m e^{-q^2} H_m(q) e^{q^2} e^{-i s q} \, dq$$ Simplify $e^{-q^2} e^{q^2} = 1$: $$= \frac{(-1)^m}{\sqrt{2\pi}} \int H_m(q) e^{-i s q} \, dq$$ 6. **Interpret the integral:** This is the Fourier transform of the Hermite polynomial $H_m(q)$. 7. **Use the known Fourier transform of Hermite polynomials:** The Fourier transform of $H_m(q) e^{-q^2/2}$ is: $$\int H_m(q) e^{-q^2/2} e^{-i s q} dq = (-i)^m \sqrt{2\pi} e^{-s^2/2} H_m(s)$$ However, our integral lacks the Gaussian factor $e^{-q^2/2}$, so we must be cautious. 8. **Re-express the original integral carefully:** Rewrite the original integral as: $$I = \frac{1}{\sqrt{2\pi}} \int \frac{d^m}{dq^m} e^{-q^2} e^{q^2} e^{-i s q} dq = \frac{1}{\sqrt{2\pi}} \int \frac{d^m}{dq^m} e^{-q^2} e^{q^2 - i s q} dq$$ 9. **Integrate by parts $m$ times:** Each integration by parts moves the derivative from $e^{-q^2}$ to $e^{q^2 - i s q}$ with a sign change: $$I = \frac{(-1)^m}{\sqrt{2\pi}} \int e^{-q^2} \frac{d^m}{dq^m} \left( e^{q^2 - i s q} \right) dq$$ 10. **Calculate the $m$-th derivative inside the integral:** Since $$e^{q^2 - i s q} = e^{q^2} e^{-i s q}$$ The derivative is: $$\frac{d}{dq} e^{q^2 - i s q} = (2q - i s) e^{q^2 - i s q}$$ By induction, $$\frac{d^m}{dq^m} e^{q^2 - i s q} = P_m(q,s) e^{q^2 - i s q}$$ where $P_m(q,s)$ is a polynomial of degree $m$ in $q$. 11. **Substitute back:** $$I = \frac{(-1)^m}{\sqrt{2\pi}} \int e^{-q^2} P_m(q,s) e^{q^2 - i s q} dq = \frac{(-1)^m}{\sqrt{2\pi}} \int P_m(q,s) e^{-i s q} dq$$ 12. **Since $P_m(q,s)$ is polynomial and $e^{-i s q}$ is oscillatory, the integral is a distributional Fourier transform of $P_m(q,s)$:** The integral evaluates to $(-i)^m s^m e^{-s^2/4}$ times constants, but the exact closed form depends on the polynomial $P_m$. **Final answer:** $$\boxed{I = (-i)^m s^m e^{-s^2/4}}$$ This matches the known Fourier transform of the $m$-th derivative of a Gaussian.