Gaussian Integral generalised

$\begin{aligned} I & = \int_{-\infty}^\infty \exp \left(-(ax^2+bx+c)\right)\ dx & \small \color{#3D99F6} \text{Note that }\exp (x) = e^x \\ & = \int_{-\infty}^\infty \exp \left(-\left(\sqrt ax+\frac b{2\sqrt a} \right)^2 + \frac {b^2}{4a} - c\right)\ dx \\ & = e^{\frac {b^2-4ac}{4a}} \int_{-\infty}^\infty e^{-\left(\sqrt ax+\frac b{2\sqrt a} \right)^2} dx & \small \color{#3D99F6} \text{Let }u = \sqrt ax+\frac b{2\sqrt a} \implies du = \sqrt a\ dx \\ & = e^{\frac {b^2-4ac}{4a}} \int_{-\infty}^\infty \frac {e^{-u^2}}{\sqrt a} du & \small \color{#3D99F6} \text{Since the integrand is even} \\ & = e^{\frac {b^2-4ac}{4a}}\cdot \frac 2{\sqrt a}\color{#3D99F6} \int_0^\infty e^{-u^2}\ du & \small \color{#3D99F6} \text{Gaussian integral }\int_0^\infty e^{-x^2}\ dx = \frac {\sqrt \pi}2 \\ & = e^{\frac {b^2-4ac}{4a}}\sqrt{ \frac \pi a} \end{aligned}$

Therefore, $\boxed{k=4, \ t=4,\ h = \pi}$ .

$I=\int_{-∞}^∞e^{-\left(ax^2+bx+c\right)}dx\$ $first\ we\ will\ write\ ax^2+bx+c\ =\ a\left(x+\frac{b}{2a}\right)^2+\frac{\left(4ac-b^2\right)}{4a}$ $So:\ I=\int_{-∞}^∞e^{-\ \left(a\left(x+\frac{b}{2a}\right)^2+\frac{\left(4ac-b^2\right)}{4a}\right)}dx=\int_{-∞}^∞e^{-\left(\ a\left(x+\frac{b}{2a}\right)^2\right)-\frac{\left(4ac-b^2\right)}{4a}}dx\$ $So:I=e^{-\frac{\left(4ac-b^2\right)}{4a}}\int_{-∞}^∞e^{-\left(\ \sqrt{a}\left(x+\frac{b}{2a}\right)\right)^2}dx\$ $now,\ let\ u=\sqrt{a}\left(x+\frac{b}{2a}\right)\ \ =>\ \ du=\sqrt{a}dx$ $So:\ I=e^{-\frac{\left(4ac-b^2\right)}{4a}}\int_{-∞}^∞\frac{e^{-\left(\ u\right)^2}du}{\sqrt{a}}\ =\frac{1}{\sqrt{a}}e^{^{\frac{\left(b^2-4ac\right)}{4a}}}\int_{-∞}^∞e^{-\left(\ u\right)^2}du$ $now\ from\ simple\ gaussian\ integral\ we\ know\ \int_{-∞}^∞e^{-u^2}du=\sqrt{\pi}$ $=>I=e^{\frac{\left(b^2-4ac\right)}{4a}}\sqrt{\frac{\pi}{a}}$ $So,\ by\ comparing\ k=t=4\ ,h=\pi$