integrating $\int\limits_{ - \infty }^\infty {\sin ax{e^{ - {x^2}}}dx} $ I want to solve the following equation:
$${u_t} = {u_{xx}} + {u_{yy}} + {e^t}$$
$$u\left( {x,y,0} \right) = \cos x\sin y$$
General form of equation:
$${u_t} = {a^2}\left( {{u_{xx}} + {u_{yy}}} \right) + f\left( {x,y,t} \right)$$
$$u\left( {x,y,0} \right) = \varphi \left( {x,y} \right)$$
The solution of the general form:
$$u\left( {x,y,t} \right) = \frac{1}{{4{a^2}\pi t}}\int\limits_{ - \infty }^\infty  {\int\limits_{ - \infty }^\infty  {\varphi \left( {\xi ,\eta } \right){e^{ - \frac{{{{\left( {x - \xi } \right)}^2} + {{\left( {y - \eta } \right)}^2}}}{{4{a^2}t}}}}} d\xi d\eta }  + \int\limits_0^t {\int\limits_{ - \infty }^\infty  {\int\limits_{ - \infty }^\infty  {\frac{{f\left( {\xi ,\eta } \right)}}{{4{a^2}\pi \left( {t - \tau } \right)}}{e^{ - \frac{{{{\left( {x - \xi } \right)}^2} + {{\left( {y - \eta } \right)}^2}}}{{4{a^2}\left( {t - \tau } \right)}}}}} d\xi d\eta } } d\tau $$
Applying the formula:
$$u\left( {x,y,t} \right) = \frac{1}{{4\pi t}}\int\limits_{ - \infty }^\infty  {\int\limits_{ - \infty }^\infty  {\cos \xi \sin \eta {e^{ - \frac{{{{\left( {x - \xi } \right)}^2} + {{\left( {y - \eta } \right)}^2}}}{{4t}}}}d\xi d\eta } }  + \frac{1}{{4\pi }}\int\limits_0^t {\frac{{{e^\tau }d\tau }}{{t - \tau }}\int\limits_{ - \infty }^\infty  {\int\limits_{ - \infty }^\infty  {{e^{ - \frac{{{{\left( {x - \xi } \right)}^2} + {{\left( {y - \eta } \right)}^2}}}{{4t}}}}d\xi } d\eta } } $$
After simplifying:
$$u\left( {x,y,t} \right) = \frac{1}{\pi }\int\limits_{ - \infty }^\infty  {\sin \left( {y + 2\sqrt t {s_1}} \right){e^{ - s_1^2}}d{s_1}} \int\limits_{ - \infty }^\infty  {\cos \left( {x + 2\sqrt t {s_2}} \right){e^{ - s_2^2}}d{s_2}}  + {e^t} - 1$$
My question is how do I integrate the last integrals?
 A: For
$$
S := \int_{-\infty}^{\infty} \sin(b+as)e^{-s^2}\,ds\qquad\text{and}\qquad
C := \int_{-\infty}^{\infty} \cos(b+as)e^{-s^2}\,ds ,
$$
we can do this: $S = \operatorname{Im} B$
and $C = \operatorname{Re} B$ where
$$
B := \int_{-\infty}^{\infty} \exp\big(-s^2+i(b+as)\big)\;ds .
$$
Complete the square
\begin{align}
B &= \int_{-\infty}^{\infty} \exp\big(-(s-\frac{ia}{2})^2+(-\frac{a^2}{4}+ib)\big)\;ds
\\&=
\exp\big(-\frac{a^2}{4}+ib\big)\int_{-\infty}^{\infty} \exp\big(-(s-\frac{ia}{2})^2\big)\;ds
\\&=
\exp\big(-\frac{a^2}{4}+ib\big)\sqrt{\pi} .
\end{align}
Thus
$$
S = \sqrt{\pi}\sin(b)\exp\big(-\frac{a^2}{4}\big)
\qquad\text{and}\qquad
C = \sqrt{\pi}\cos(b)\exp\big(-\frac{a^2}{4}\big)
$$
A: We try to evaluate $$f(\alpha,\beta)=\int_{-\infty}^\infty e^{-x^2}\cos(\alpha x+\beta)dx.$$
Note that
$$
f(\alpha,\beta){=\int_{-\infty}^\infty e^{-x^2}\cos(\alpha x+\beta)dx
\\=
\int_{-\infty}^\infty e^{-x^2}[\cos\alpha x\cos\beta-\sin\alpha x\sin\beta] dx
\\=
\cos\beta\int_{-\infty}^\infty e^{-x^2}\cos\alpha x dx
.
}
$$
Using integration by parts for $g(\alpha)=\int_{-\infty}^\infty e^{-x^2}\cos\alpha x dx$, we obtain
$$
\int_{-\infty}^\infty e^{-x^2}\cos\alpha x dx{=
\frac{1}{\alpha}e^{-x^2}\sin\alpha x\Big|_{-\infty}^\infty+\frac{2}{\alpha}\int_{-\infty}^\infty xe^{-x^2}\sin\alpha x dx
\\=
\frac{2}{\alpha}\int_{-\infty}^\infty xe^{-x^2}\sin\alpha x dx
\\=
-\frac{2}{\alpha}g'(\alpha).
}
$$
Hence
$$
\frac{g'(\alpha)}{g(\alpha)}=-\frac{\alpha}{2}\implies g(\alpha)=Ce^{-\frac{\alpha^2}{4}},
$$
where $C=g(0)=\int_{-\infty}^\infty e^{-x^2}dx=\sqrt\pi$. Finally,
$$
\int_{-\infty}^\infty e^{-x^2}\cos(\alpha x+\beta)dx=\sqrt\pi e^{-\frac{\alpha^2}{4}}\cos\beta.
$$
A: Compute instead
$$I=\int e^{-s^2+i \left(2 s \sqrt{t}+y\right)}\,ds$$ Complete the square to face a well know integrals, making
$$I=\frac{\sqrt{\pi }}{2}\,  e^{-t+i y}\, \text{erf}\left(s-i
   \sqrt{t}\right)$$
The question now is : do the real and imaginary parts converge for infinite bounds ?
