TITLE: Show that $\int_0^\infty e^{-x^2} \sin{2xb}\, dx =e^{-b^2}\int_0^b e^{x^2} \, dx, \, b>0, $?
QUESTION [6 upvotes]: Show that $$\int_0^\infty e^{-x^2} \sin{2xb}\, dx =e^{-b^2}\int_0^b e^{x^2} \, dx, \, b>0, $$
I need help. I did the following steps:
Apply Cauchy's Theorem, being $\varphi (x) = e^{-z^2}$ analytic in whole complex plane; take rectangle $0 \leq x \le a, \, 0\le y\le b$. Then,
$$ 0=\int_0^a e^{-x^2} \, dx + \int_0^b e^{-(a+iy)^2} i \, dy + \int_a^0 e^{-(x+ib)^2} \, dx + \int_b^0 e^{-(-a+iy)^2} i \, dy$$
$$= \int_0^a e^{-x^2} \, dx - \int_0^b e^{(-a^2-y^2-2ayi)} i \, dy +  \int_a^0 e^{(-x^2-2xbi+b^2)} \, dx +  \int_b^0 e^{(-a^2+y^2+2ayi)}i \, dy $$
$$ =  \int_0^a e^{-x^2} \, dx -e^{b^2} \int_0^a e^{-x^2} (\cos{(-2bx)}+i\sin{(-2bx)}) \, dx + \left[  -ie^{-a^2}\int_0^b e^{y^2} (e^{2iay}- e^{-2iay} \, dy \right]$$
$$ =  \int_0^a e^{-x^2} \, dx -e^{b^2} \int_0^a e^{-x^2} (\cos{(2bx)}-i\sin{(2bx)}) \, dx + \left[  -ie^{-a^2}\int_0^b e^{y^2} (e^{2iay}- e^{-2iay}) \, dy \right]$$
$$= \int_0^a e^{-x^2} \, dx -e^{b^2} \int_0^a e^{-x^2} (\cos{(2bx)}-i\sin{(2bx)}) \, dx + 2e^{-a^2}\int_0^b e^{y^2}\sin(2ay) \, dy  $$
I`m doing right? Any suggestions? I think that the question need to be reformulated.

REPLY [2 votes]: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
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 \newcommand{\pars}[1]{\left( #1 \right)}%
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\begin{align}
\int_{0}^{\infty}\expo{-x^{2}}\sin\pars{2xb}\,\dd x
&=
\Im\int_{0}^{\infty}\expo{-\pars{x^{2} - 2\ic bx}}\,\dd x
=
\Im\int_{0}^{\infty}\expo{-\pars{x - \ic b}^{2} - b^{2}}\,\dd x
\\[3mm]&=
\expo{-b^{2}}\Im\int_{-\ic b}^{\infty - \ic b}\expo{-x^{2}}\,\dd x
\\[3mm]&=
\expo{-b^{2}}\Im\bracks{%
\int_{0}^{\infty}\expo{-x^{2}}\,\dd x + \int_{-b}^{o}\expo{-\pars{\ic x}^{2}}\,
\pars{\ic \dd x}}
=
\expo{-b^{2}}\int_{-b}^{0}\expo{x^{2}}\,\dd x
\\[5mm]&
\end{align}
$${\large%
\int_{0}^{\infty}\expo{-x^{2}}\sin\pars{2xb}\,\dd x
=
\expo{-b^{2}}\int_{0}^{b}\expo{x^{2}}\,\dd x}
$$