Show that $\int_0^\infty e^{-x^2} \sin{2xb}\, dx =e^{-b^2}\int_0^b e^{x^2} \, dx, \, b>0, $? 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.
 A: Right idea - wrong rectangle.  You need an infinite rectangle, or at least the limit of one - because you want an improper integral.  Thus, I would use the rectangle with vertices $-i b$, $R-i b$, $R+i b$, and $i b$; take the limit as $R \to \infty$.  Thus we have
$$\oint_C dz \, e^{-z^2} = 0$$
by Cauchy's theorem.  Now let's write out that contour integral:
$$\int_0^R dx \, e^{-(x-i b)^2} + i \int_{-b}^b dy \, e^{-(R+i y)^2} \\ + \int_R^0 dx \, e^{-(x+i b)^2} + i \int_b^{-b} dy \, e^{y^2} = 0$$
Note that the second integral vanishes as $R \to \infty$.  Rearranging things a bit, I get
$$e^{b^2} \int_0^{\infty} dx \, e^{-x^2} \, e^{i 2 b x} - e^{b^2}\int_0^{\infty} dx \, e^{-x^2} \, e^{-i 2 b x} - i \int_{-b}^b dy \, e^{y^2} = 0$$
from which I get the desired result:
$$\int_0^{\infty} dx \, e^{-x^2} \, \sin{2 b x} = e^{-b^2} \int_0^b dy \, e^{y^2}$$
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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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}
$${\large%
\int_{0}^{\infty}\expo{-x^{2}}\sin\pars{2xb}\,\dd x
=
\expo{-b^{2}}\int_{0}^{b}\expo{x^{2}}\,\dd x}
$$
A: Setting
$$
f(t)=e^{t^2}\int_0^\infty e^{-x^2}\sin(2tx)\,dx,
$$
we have for every $t>0$:
\begin{eqnarray}
f'(t)&=&2tf(t)+e^{t^2}\int_0^\infty 2xe^{-x^2}\cos(2tx)\,dx\\
&=&2tf(t)-e^{t^2}\left[e^{-x^2}\cos(2tx)\right]_0^\infty-2te^{t^2}\int_0^\infty e^{-x^2}\sin(2tx)\,dx\\
&=&2tf(t)+e^{t^2}-2tf(t)=e^{t^2}.
\end{eqnarray}
Thus, for every $0<\varepsilon<b$ we have
$$
e^{-b^2}f(b)=e^{-b^2}\left[f(\varepsilon)+\int_\varepsilon^b e^{x^2}\,dx\right].
$$
Letting $\varepsilon \downarrow 0$ we get:
$$
\int_0^\infty e^{-x^2}\sin(2bx)\,dx=e^{-b^2}f(b)=e^{-b^2}\int_0^b e^{x^2}\,dx.
$$
