Let $b>0$. Show that $\int_{-\infty}^{+\infty}e^{-t^2}\cos(2b\pi t)dt=\sqrt{\pi}e^{-b^2\pi^2}$ Let $b>0$. Show that $\int_{-\infty}^{+\infty}e^{-t^2}\cos(2b\pi t)dt=\sqrt{\pi}e^{-b^2\pi^2}$
The book gives a help to solve this exercise but even so I am not able to solve it, the help is the following:
(Consider $\int_{\partial R}e^{-z^2}dz$, where $R$ is the rectangule with vertices $-c, c, c+b\pi i$ and $-c+b\pi i$ for $c>0$. Recall Exercise IV$.4.18.$)
Exercise IV$.4.18.$ Let $a$ and $b$ be real numbers satisfying $a<b$, and let $I(c)$ be defined for any real number $c$ by $I(c)=\int_{c+ia}^{c+ib}e^{-z^2}dz$. By deriving a suitable upper bound for $|I(c)|$ conclude that $I(c)\rightarrow 0$ as $c\rightarrow \pm \infty$
And here is something interesting about this: Bounds for $I(c)=\int_{c+ia}^{c+ib}e^{-z^2}\,dz$.
Could someone explain to me why I should use exercise 4.18? Why should I first consider that integral in that rectangle? That integral would not give zero for Cauchy's theorem? Thank you very much.

$\displaystyle 0 = \int_{\partial R}e^{-z^2}dz = \int_{[-c, \,c]}e^{-z^2}dz + \int_{[c, \, c + b\pi i]}e^{-z^2}dz + \int_{[c + b\pi i, \, -c + b\pi i]}e^{-z^2}dz + \int_{[-c + b\pi i, \, -c]}e^{-z^2}dz$
$\displaystyle = \int_{-c}^{c} e^{-x^2}dx + \int_{c}^{c+ i b\pi} e^{-z^2}dz + \int_{c}^{-c} e^{-(t+ib\pi)^2}dt + \int_{-c+ i b\pi}^{-c} e^{-z^2}dz$
$\displaystyle = \int_{-c}^{c} e^{-x^2}dx + \int_{c}^{c+ i b\pi} e^{-z^2}dz + \int_{c}^{-c} e^{-t^2+b^2\pi^2 - i2b\pi t}dt + \int_{-c+ i b\pi}^{-c} e^{-z^2}dz$
$\displaystyle = \int_{-c}^{c} e^{-x^2}dx + \int_{c}^{c+ i b\pi} e^{-z^2}dz + \int_{c}^{-c} e^{-t^2+b^2\pi^2}\cos(2\pi b t)dt - i \int_{c}^{-c} e^{-t^2+b^2\pi^2}\sin(2\pi b t)dt + \int_{-c+ i b\pi}^{-c} e^{-z^2}dz$
Now letting $c \to \infty$ and using your Exercise IV 4.18 we get
$\displaystyle 0 = \int_{-\infty}^{\infty} e^{-x^2}dx + \int_{\infty}^{-\infty} e^{-t^2+b^2\pi^2}\cos(2\pi b t)dt - i \int_{\infty}^{-\infty} e^{-t^2+b^2\pi^2}\sin(2\pi b t)dt \quad$ that is
$\displaystyle 0 = \sqrt{\pi} - e^{b^2 \pi^2} \int_{-\infty}^{\infty}e^{-t^2}\cos(2b\pi t) dt - i e^{b^2 \pi^2} \int_{-\infty}^{\infty}e^{-t^2}\sin(2b\pi t) dt \quad$ So we find
$\displaystyle \int_{-\infty}^{\infty}e^{-t^2}\cos(2b\pi t) dt = \sqrt{\pi}e^{-b^2\pi^2}$ and $\displaystyle \int_{-\infty}^{\infty}e^{-t^2}\sin(2b\pi t) dt = 0$. 
Note: The first line comes from analyticity of $e^{-z^2}$ on and in the rectangle. That is, integral of an analytic function on a simple closed curve is $0$.
The fourth line comes from $\cos(-2b\pi t) = \cos(2b\pi t)$ and $\sin(-2b\pi t) = -\sin(2b\pi t)$, that is evenness and oddness of cosine and sine respectively.
The fifth line comes from the well-known Gaussian integral $\displaystyle \int_{-\infty}^{\infty}e^{-x^2} = \sqrt{\pi}$ and reversing limits of the integral, that is $\displaystyle \int_a^b f(t) dt = -\int_b^a f(t) dt \,$ i.e. $\, \displaystyle \int_{\infty}^{-\infty} f(t) dt = -\int_{-\infty}^{\infty} f(t) dt$.
