I was going through the calculation of the free-particle kernel in Feynman and Hibbs (pp 43). The book describes $$ \left(\frac{m}{2\pi i\hbar\epsilon}\right)\int_{-\infty}^{\infty}\exp\left(\frac{im}{2\hbar\epsilon}[(x_2-x_1)^2+(x_1-x_0)^2]\right)dx_1 \tag{3.4} $$ as a Gaussian integral. And the result of this integration is given as $$ \left(\frac{m}{2\pi i\hbar \cdot 2\epsilon}\right)^{1/2}\exp\left(\frac{im}{2\hbar\cdot 2\epsilon}(x_2-x_0)^2\right). \tag{3.4} $$ The only way I could do this integral was treating it as a Gaussian integral over $ix_1$, in which case I get the required solution but with a negative sign upfront. Is this the right way to do it, or is there something very obvious that I am not seeing?
On a related note, problem 3-8 (pp 63) also involves a similar "Gaussian" integral in order to obtain the kernel for the quantum harmonic oscillator. My technique of performing a Gaussian integral over $ix$ goes horribly wrong there and I am not even near the correct answer which should be of the form $$ \frac{1}{\sqrt{2\pi\sin\epsilon}}\exp\left(\frac{i}{2\sin\epsilon}[(x_0^2+x_2^2)\cos\epsilon-2x_2x_2]\right) $$ after performing the integration, $$ \exp\left(\frac{i\cot\epsilon}{2}(x_0^2+x_2^2)\right)$$ $$\times \int\exp\left(\frac{i}{2\sin\epsilon}[2x_1^2\cos\epsilon-2x_1(x_0+x_2)]\right)dx_1. $$ Any help will be greatly appreciated.