Help in solving an integral. I am trying to evaluate this integral, but could not find a solution. I tried it, assuming it to be product of two exponential and then tried integration by parts but it does not lead to anywhere. Can anyone please help?
$$ \int_{-T}^{T} \; e^{(f-ct)x+kx^2} \; dx $$
It seems the integral is not possible as per comment. So can we evaluate the integral using principle of stationary phase (assuming the limit of integration is infinity) and then somehow estimate for bounded T?
 A: In the case where $k > 0$, let's let $a = \sqrt{k}$, so we have
\begin{align}
\int_{-T}^{T} \; e^{(f-ct)x+kx^2} \; dx
&= \int_{-T}^{T} \; \exp({(f-ct)x+a^2x^2} )\; dx\\
&= \int_{-T}^{T} \; \exp({(ax + \frac{f-ct}{2a})^2 - \left[ \frac{f-ct}{2a} \right]^2 }) \; dx\\
&= \int_{-T}^{T} \; \exp({(ax + \frac{f-ct}{2a})^2) \exp(- \left[ \frac{f-ct}{2a} \right]^2 }) \; dx\\
&= \exp(- \left[ \frac{f-ct}{2a} \right]^2 ) \int_{-T}^{T} \; \exp({(ax + \frac{f-ct}{2a})^2) } \; dx
\end{align}
Let's now call that thing in front $C$, so that we have
\begin{align}
\int_{-T}^{T} \; e^{(f-ct)x+kx^2} \; dx
&= K \int_{-T}^{T} \; \exp({(ax + \frac{f-ct}{2a})^2) } \; dx
\end{align}
Letting 
$$
u = ax + \frac{f-ct}{2a} \\
du = 2a ~ dx \\
dx = \frac{1}{2a} ~du
$$
so the integral becomes
\begin{align}
\int_{-T}^{T} \; e^{(f-ct)x+kx^2} \; dx
&= K \int_{-T}^{T} \; \exp({(ax + \frac{f-ct}{2a})^2) } \; dx\\
&= \frac{K}{2a} \int_{x=-T}^{x=T} \; \exp(u^2)  \; du
\end{align}
...and that's about as much as you're going to be able to simplify it. (I didn't transform the limits of integration, because that would just have made it messier...)
On the good side, if $T$ is growing larger and larger, the integral's headed to $\infty$, and headed there fast. 
For the case $k < 0$, write $a = \sqrt{-k}$, and do something similar, and you'll end up with an integral of $\exp(-u^2)$, which is much more well-behaved in the limit, but more or less hopeless to integrate in elementary term for finite limits. :)
A: Assume $k<0$, otherwise the integral diverges for infinite $T$.
We use a linear change of variable to normalize the exponent, with
$$(f-ct)x+kx^2=k(x+\frac{f-ct}{2k})^2-\frac{(f-ct)^2}{4k}=-\left(\sqrt{|k|}x+\frac{f-ct}{2\sqrt{|k|}}\right)^2-\frac{(f-ct)^2}{4k}.$$
Then
$$\int_{-T}^T\exp((f-ct)x+kx^2)\,dx=\exp\left(-\frac{(f-ct)^2}{4k}\right)\frac1{\sqrt{|k|}}\int_{-\sqrt{|k|}T+\frac{f-ct}{2\sqrt{|k|}}}^{\sqrt{|k|}T+\frac{f-ct}{2\sqrt{|k|}}}\exp(-u^2)\,du\\
=\frac{\sqrt\pi}2\exp\left(-\frac{(f-ct)^2}{4k}\right)\frac1{\sqrt{|k|}}\left(\text{erf}\left(\sqrt{|k|}T+\frac{f-ct}{2\sqrt{|k|}}\right)-\text{erf}\left(-\sqrt{|k|}T+\frac{f-ct}{2\sqrt{|k|}}\right)\right).$$
When $T$ goes to infinity, the difference of $\text{erf}$ quickly tends to $2$.
