In this paper (page 7), the following integral is to be solved:
$$A(x,t) = \frac{\sigma_x}{2 \sqrt{\pi}} \int_{-\infty}^{+\infty} e^{i kx} e^{- i \left[ k_c v_p + (k - k_c)v_g + \frac{1}{2}(k - k_c)^2 \Gamma \right] t} e^{- \frac{\sigma_x^2}{2}(k - k_c)^2} e^{i x_0 (k - k_c)} dk$$
where $i$ is the imaginary unit and $\sigma_x, k_c, v_p, v_g, \Gamma, x_0 \in \mathbb{R}$ are constants; $x, k, t \in \mathbb{R}$ are variables.
The solution should be
$$A(x,t) = \exp \left[ \frac{1}{2} \left( \frac{x - (x_0 + v_gt)}{\sqrt{\sigma_x^2 - i \Gamma t}} \right)^2 \right] e^{i k_c x} e^{- ik_c t(v_g - v_p)}$$
But my result is different. From the initial integral, I get
$$A(x,t) = \frac{\sigma_x}{2 \sqrt{\pi}} \int_{-\infty}^{+\infty} e^{-i k_c v_p t} e^{i k_c v_g t} e^{-i x_0 k_c} e^{i \left[ kx - \left( k v_g + \frac{1}{2}(k - k_c)^2 \Gamma \right) \right] t} e^{- \frac{\sigma_x^2}{2}(k - k_c)^2} e^{i x_0 k} dk =$$
$$ = \frac{\sigma_x}{2 \sqrt{\pi}} e^{-i k_c v_p t} e^{i k_c v_g t} e^{-i x_0 k_c} \int_{-\infty}^{+\infty} e^{i (x - v_g t + x_0) k} e^{- (k - k_c)^2 \left( i \frac{1}{2} \Gamma t + \frac{\sigma_x^2}{2} \right)} dk$$
The integral is in the Gaussian generalized form
$$\int_{-\infty}^{+\infty} e^{-\beta (x - \alpha)^2} e^{\gamma x} dx = \sqrt{\frac{\pi}{\beta}} e^{\frac{\gamma^2}{4\beta} + \alpha \gamma}$$
So,
$$A(x,t) = \frac{\sigma_x}{2 \sqrt{\pi}} e^{-i k_c v_p t} e^{i k_c v_g t} e^{-i x_0 k_c} \left[ \sqrt{\frac{\pi}{i \frac{1}{2} \Gamma t + \frac{\sigma_x^2}{2}}} e^{- \frac{(x - v_g t + x_0)^2}{2i \Gamma t + 2 \sigma_x^2}} e^{i k_c (x - v_g t + x_0)} \right]$$
This is far from being the expected result, where the square root doesn't even exist.
Did I make some (algebraic) mistake?
Is there a way (maybe transforming the square root in complex exponential?) to arrange my solution to fit to the provided solution above?
There is a problem in the integral as is written as $kx$ should not be multiplied by time $t$. This is a straightforward computation of quantum mechanics. So, I rewrite the integral as (I have also changed $x_0\rightarrow -x_0$) $$ A(x,t) = \frac{\sigma_x}{2 \sqrt{\pi}} \int_{-\infty}^{+\infty} e^{i kx - i\left[ k_c v_p + (k - k_c)v_g + \frac{1}{2}(k - k_c)^2 \Gamma \right] t} e^{- \frac{\sigma_x^2}{2}(k - k_c)^2} e^{-i x_0 (k - k_c)} dk. $$ It is useful at this stage to introduce a new integration variable $\tilde k=k-k_c$. Then one has $$ A(x,t) = \frac{\sigma_x}{2 \sqrt{\pi}}e^{ik_cx}e^{-ik_cv_pt} \int_{-\infty}^{+\infty} e^{i{\tilde k}x - i\left[ {\tilde k}v_g + \frac{1}{2}{\tilde k}^2 \Gamma \right] t} e^{- \frac{\sigma_x^2}{2}{\tilde k}^2} e^{-i{\tilde k} x_0} d{\tilde k}. $$ This is $$ A(x,t) = \frac{\sigma_x}{2 \sqrt{\pi}}e^{ik_cx}e^{-ik_cv_pt} \int_{-\infty}^{+\infty}e^{-\frac{1}{2}\left[\sigma_x^2+i\Gamma t\right]{\tilde k}^2} e^{i{\tilde k}(x-x_0-v_gt)}d{\tilde k}. $$ This integral can be evaluated using the formula $$ \int_{-\infty}^{+\infty}e^{-\frac{\alpha}{2}x^2+\beta x}dx=\sqrt{\frac{2\pi}{\alpha}}e^{\frac{\beta^2}{2\alpha}} $$ that yields the final result $$ A(x,t)=\frac{\sigma_x}{2 \sqrt{\pi}}e^{ik_cx}e^{-ik_cv_pt}\sqrt{\frac{2\pi}{\sigma_x^2+i\Gamma t}}e^{-\frac{(x-x_0-v_gt)^2}{2\left(\sigma_x^2+i\Gamma t\right)}}. $$
I guess that the author of the lecture is mostly interested in the spreading of the wave-packet in time rather than to obtain the exact formula with all the constants right in place.