Fourier transforms of $f(t)=\frac{\sin{at}}{t}$ 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.
 A: 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!
A: There is another easy way of calculating the Fourier Transform instead of going the direct route.
In order to use that route, you will need to understand the concept of duality. The duality is a concept relating $t$ to $\lambda$. Basically, if $F(\lambda)$ is the Fourier Transform of $f(t)$, then $2\pi f(-\lambda)$ is the Fourier Transform of $F(t)$.
If you go to this link http://web.mit.edu/shou/www/6.003/tutorial_tables.pdf
you will find what you are looking for in page 4.
