I want to derive the following pair of Fourier transforms:
First: $$f(t)=\dfrac{\sin{at}}{t}$$
$$F(\lambda)= \begin{cases} \sqrt{\dfrac{\pi}{2}}, & \text{if } |\lambda|<a \\ 0, & \text{if } |\lambda|>a \end{cases}$$
Second:
$$f(t)=(a^2+t^2)^{-1}$$
$$F(\lambda)=\sqrt{\dfrac{\pi}{2}}\cdot\dfrac{e^{-a|\lambda|}}{a}$$
What I have done: I have started using the definition of $F(\lambda)$, the fourier transform of $f(t)$, as: $$F(\lambda)=\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{e^{i\lambda \tau}f(\tau)d\tau}$$ So, for the first function: $f(t)=\dfrac{\sin{at}}{t}$ $$\implies F(\lambda)=\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{e^{i\lambda \tau}\dfrac{\sin{a\tau}}{\tau}d\tau}$$
But I don't know how to integrate this.
METHOD 1:
There are several ways to go here. If we are permitted use of Generalized Functions, then we can proceed as follows. Let $I(\lambda)$ be the integral
$$I(\lambda)=\int_{-\infty}^{\infty}e^{i\lambda\tau}\frac{\sin a\tau}{\tau}d\tau\tag 1$$
Taking the derivative with respect to $\lambda$ of $(1)$ gives
$$\begin{align} I'(\lambda)&=i\int_{-\infty}^{\infty}e^{i\lambda\tau}\sin a\tau\,d\tau\\\\ &=i\int_{-\infty}^{\infty}e^{i\lambda\tau}\frac{e^{i a\tau}-e^{-ia\tau}}{2i}\,d\tau\\\\ &=\pi \left(\delta(\lambda+a)-\delta(\lambda-a)\right)\tag 2 \end{align}$$
where $\delta$ is the Dirac Delta. Integrating $(2)$ reveals that
$$I(\lambda)=\pi \left(H(\lambda+a)-H(\lambda-a)\right)+C$$
where $H$ is the Heaviside Function and $C$ is a constant of integration. For $a=0$, $I=0$, which implies that $C=0$. We have finally the coveted result
$$\bbox[5px,border:2px solid #C0A000]{\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{i\lambda \tau}\frac{\sin a \tau}{\tau}\,d\tau=\sqrt{\frac{\pi}{2}}\left(H(\lambda+a)-H(\lambda-a)\right)}$$
METHOD 2:
A second approach uses classical analysis only and forgoes the use of Generalized Functions.
Here, we will find the Fourier Transform of $\frac{\sin at}{t}$ by first finding its Fourier Transform representation and subsequently invoking the Fourier Inversion Theorem.
To that end, we have
$$\begin{align} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\sqrt{\frac{\pi}{2}}\left(H(\lambda+a)-H(\lambda-a)\right)\,e^{-i\lambda t}d\lambda&=\frac12 \int_{-a}^{a}e^{-i\lambda t}d\lambda\\\\ &=\frac12\left(\frac{e^{iat}-e^{-iat}}{it}\right)\\\\ &=\frac{\sin at}{t} \end{align}$$
Finally, using the Fourier Inversion Theorem, we have
$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\left(\frac{\sin at}{t}\right)(\lambda)=\sqrt{\frac{\pi}{2}}\left(H(\lambda+a)-H(\lambda-a)\right)}$$
METHOD 3:
Here, we use contour integration to obtain the result. We write
$$\int_{-\infty}^{\infty}e^{i\lambda\tau}\frac{\sin a\tau}{\tau}d\tau=\int_{-\infty}^{\infty}\frac{e^{i(\lambda+a)\tau}-e^{-i(\lambda+a)\tau}}{2i\tau}d\tau$$
We evaluate the integral
$$J(k)=\oint_C\frac{e^{ik\tau}}{\tau}d\tau$$
in the complex $\tau$-plane, where for $k>0$ ($k<0$), $C$ is the contour comprised of
(i) the real-line segment from $(-R,0)$ to $(-\epsilon,0)$ $\epsilon>0$,
(ii) the real-line from $(\epsilon,0)$ to $(R,0)$,
(iii) the semi-circle $C_{R^{+}}$ ($C_{R^{-}}$) in the upper-half (lower-half )plane with radius $R$ and centered at $(0,0)$ from $(R,0)$ to $(-R,0)$, and
(iv) the semi-circle $C_{\epsilon^{+}}$ ($C_{\epsilon^{-}}$) in the upper half (lower-half) plane with radius $\epsilon$ centered at $(0,0)$ from $(-\epsilon,0)$ to $(\epsilon,0)$.
First, we see from the Residue Theorem that $J=0$.
Second, as $R\to \infty$, we can use Jordan's Lemma to show that the contribution from the integral over $C_{R^{+}}$ ($C_{R^{-}}$) vanishes.
Next, we examine the integral around $C_{\epsilon^{+}}$ ($C_{\epsilon^{-}}$) in the limit as $\epsilon \to 0$. We have
$$ \lim_{\epsilon \to 0}\int_{C_{\epsilon^{\pm}}}\frac{e^{ik\tau}}{\tau}d\tau= \begin{cases} -i\pi i ,&k>0\\\\ +i\pi i ,&k<0 \end{cases}$$
Thus, we are left with the Cauchy Principal Value integral
$$\int_{-\infty}^{\infty}\frac{e^{ik\tau}}{\tau}d\tau=\lim_{\epsilon \to 0,R\to \infty}\left(\int_{-R}^{-\epsilon}\frac{e^{ik\tau}}{\tau}d\tau+\int_{\epsilon}^{R}\frac{e^{ik\tau}}{\tau}d\tau\right)=i\pi\,\text{sgn}(k)$$
where $\text{sgn}$ is the Sign Function. Therefore, we have
$$\begin{align} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{i\lambda\tau}\frac{\sin a\tau}{\tau}d\tau&=\frac{1}{\sqrt{2\pi}}i\pi\left(\frac{\text{sgn}(\lambda+a)-\text{sgn}(\lambda-a)}{2i}\right)\\\\ &=\sqrt{\frac{\pi}{2}}\left(H(\lambda+a)-H(\lambda-a)\right) \end{align}$$
as expected!