Evaluate $\int_{-\infty}^{\infty} e^{-(2t+1)^2} e^{-ixt}\,dt$ I am trying to evaluate:
$$\int_{-\infty}^\infty e^{-(2t+1)^2} e^{-ixt}\,dt$$
then
$$\int_{-\infty}^\infty e^{-(2t+1)^2} [\cos(xt)-i\sin(xt)]\,dt$$
since $\sin(xt)$ is odd, it remains:
$$\int_{-\infty}^\infty e^{-(2t+1)^2} \cos(xt)\,dt$$
So I couldn't integrate this last integral.
 A: \begin{align}
& (2t+1)^2 + ixt = 4t^2+4t+1+ixt = 4t^2 + (4+ix)t + 1 \\[10pt]
= {} & 4 \left( t^2 + \left( 1 + \frac{ix} 4 \right)t\right) + 1 \\[10pt]
= {} & 4 \underbrace{\left( t^2 + \left( 1 + \frac{ix} 4 \right)t + \left( \frac 1 2 + \frac{ix} 8 \right)^2 \right)}_\text{This is the complete square.} + 1 - 4\left( \frac 1 2 + \frac{ix} 8 \right)^2 \\[10pt]
= {} & 4 \overbrace{\left( t + \frac 1 2 + \frac{ix} 8 \right)^2}^\text{So is this.}  \underbrace{{} - \frac{ix} 2 + \frac{x^2}{16}}_{ \begin{smallmatrix} \text{The variable $t$ does} \\  \text{not appear in these} \\  \text{three terms.} \end{smallmatrix} } \\[10pt]
= {} & \left( 2t + 1 + \frac{ix} 4 \right)^2 - \frac{ix} 2 + \frac{x^2}{16}.
\end{align}
So the integral is
$$
e^{\left( \displaystyle - \frac{ix} 2 + \frac{x^2}{16} \right)}  \int_{-\infty}^\infty e^{\displaystyle-\left( 2t+1+ \frac{ix} 4 \right)^2} \, dt. \tag 0
$$
If you can show this last integral (without the factor that was pulled out in front) is equal to
$$
\int_{-\infty}^\infty e^{-4t^2} \, dt,
$$
then we're almost done.
$$
\int_{-\infty}^\infty e^{-4t^2} \, dt = \int_{-\infty}^\infty e^{-\frac{t^2}{2\sigma^2}} \, dt \quad\text{where }\sigma = \frac 1 {2\sqrt 2}. \tag 1
$$
The usual form of the normal density tells us that
$$
\frac 1 {\sigma\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac{t^2}{2\sigma^2}} \, dt = 1,
$$
so that tells you the value of $(1)$.
But why should
$$
\int_{-\infty}^\infty e^{\displaystyle-\left( 2t+1+ \frac{ix} 4 \right)^2} \, dt
$$
be equal to
$$
\int_{-\infty}^\infty e^{-4t^2} \, dt \text{ ?}
$$
One could try to do this by substitution: $2u = 2t + 1 + \dfrac{ix} 2 $, $du = dt$, etc.  Then as $t$ goes from $-\infty$ to $+\infty$, then $u$ goes from what to what?  From $-\infty+\text{some imaginary number}$ to $+\infty+\text{the same imaginary number}$.  How to deal with the imaginary part?
Consider
$$
\int_{-M}^M e^{-2t^2} \,dt + \int_M^{M+ ia} \cdots + \int_{M+ia}^{-M+ia} \cdots + \int_{-M+ia}^{-M} \cdots
$$
following straight lines from the lower to the upper bound of integration in each case.  This is an integral of an entire function over a closed path and is therefore $0$.
If we can now show that the two parts
$$
\int_M^{M+ia} e^{-2t^2}\,dt \text{ and } \int_{-M+ia}^{-M} \cdots \tag 2
$$
approach $0$ as $M\to\infty$, then the two integrals from $-\infty$ to $+\infty$ and from $-\infty+ia$ to $+\infty+ia$ must both be the same.
The two limits in $(2)$ are found by getting a bound on the size of the function on those two paths.
A: $\begin{eqnarray}
\int_{-\infty}^\infty e^{-(2t+1)^2}e^{-itx}dt&=&\int_{-\infty}^\infty e^{-(1+4t+4t^2)}e^{-itx}dt\\
&=&\int_{-\infty}^\infty e^{-4(t^2+t+\frac{1}{4})}e^{-itx}dt\\
&=&\int_{-\infty}^\infty e^{-\frac{\left(t+\frac{1}{2}\right)^2}{1/4}}e^{-itx}dt\\
&=&\frac{\sqrt{2\pi}}{\sqrt{8}}\int_{-\infty}^\infty \frac{\sqrt{8}}{\sqrt{2\pi}}e^{-\frac{\left(t+\frac{1}{2}\right)^2}{2\frac{1}{8}}}e^{-itx}dt\\
&=&\frac{\sqrt{2\pi}}{\sqrt{8}}M_T(-ix)\tag1
\end{eqnarray}$
where $T$ is a normal with $\mu=-\frac{1}{2}$ and $\sigma^2=\frac{1}{8}$ and $M_T$ is its generating moments function.
Thus, in $(1)$, $M_T(-ix)=e^{\mu (-ix)+\frac{\sigma^2(ix)^2}{2}}=e^{-\frac{ix}{2}-\frac{x^2}{16}}$.
Therefore $$\int_{-\infty}^\infty e^{-(2t+1)^2}e^{-itx}dt=\frac{\sqrt{2\pi}}{\sqrt{8}}e^{-\frac{ix}{2}-\frac{x^2}{16}}$$
