Inverse Fourier Transform of a squared sinc function I have here a squared sinc function, which is the Fourier Transform of some triangular pulse:
$$\mathrm H(f)= 2\mathrm A\mathrm T_\mathrm o \frac{\sin^2(2\pi f \mathrm T_\mathrm o)}{(2\pi f\mathrm T_\mathrm o)^2}$$
As an excercise, I would like to go back to the original time domain triangular pulse, using the inverse Fourier Transform.
I feel like I'm very close to achieving it, however, I stumbled upon this integral:
$$\mathrm h(t)=\frac {\mathrm A} {2\pi ^2 \mathrm T_\mathrm o}\int_{0}^{\infty}\frac{(1-\cos(4\pi f\mathrm T_\mathrm o))\cos(2\pi ft)}{f^2}df$$
I'm not sure if there is already an expression which defines the value of this integral, or if this integral is of a well-known type, with a well-defined value. The tables of integrals that I have right now are basic and don't cover integrals of such type.
Does someone know if there is a well-known expression for the value of integrals of this type?
Or, would you be so kind as to recommend a good table of integrals for Fourier Analysis?
Thank you very much!
For $T_o&gt;0$, this integral can be tackled with a Schwinger parameterization and Fubini's Theorem.
(The equivalent of the intermediate integral you specifically asked about is at the 7th equal sign below, with the trig functions written in terms of complex exponentials.)
$$\begin{align*}h(t) &amp;= 2AT_o\int_{-\infty}^{\infty} \frac{\sin^2(2\pi fT_o)}{(2\pi fT_o)^2}e^{2\pi ift} \; df\\
\\
&amp;= 2AT_o\int_{-\infty}^{\infty} \frac{\sin^2(2\pi fT_o)}{(2\pi fT_o)^2}\cos(2\pi ft) \; df\\
\\
&amp;= 4AT_o\int_{0
}^{\infty} \frac{\sin^2(2\pi fT_o)}{(2\pi fT_o)^2}\cos(2\pi ft) \; df\\
\\
&amp;= 4AT_o\int_{0}^{\infty} \sin^2(2\pi fT_o)\cos(2\pi  ft)\int_0^{\infty}ue^{-u2\pi fT_o} \; du\; df \\
\\
&amp;= 4AT_o\int_{0}^{\infty} \int_{0}^{\infty} u\sin^2(2\pi fT_o)\cos(2\pi  ft)e^{-u2\pi fT_o} \; df\; du \\
\\
&amp;= -\dfrac{AT_o}{2}\int_{0}^{\infty} \int_{0}^{\infty} u\left(e^{i2\pi fT_o}-e^{-i2\pi fT_o}\right)^2\left(e^{i2\pi  ft}+e^{-i2\pi  ft}\right)e^{-u2\pi fT_o} \; df\; du \\
\\
&amp;= -\dfrac{AT_o}{2}\int_{0}^{\infty} \int_{0}^{\infty} u\left(e^{i2\pi f 2T_o}-2+e^{-i2\pi f 2T_o}\right)\left(e^{i2\pi  ft}+e^{-i2\pi  ft}\right)e^{-u2\pi fT_o} \; df\; du \\
\\
&amp;= -\dfrac{AT_o}{2}\int_{0}^{\infty} \int_{0}^{\infty} u\left(e^{i2\pi f (2T_o+t)}+e^{i2\pi f (2T_o-t)}-2\left[e^{i2\pi  ft}+e^{-i2\pi  ft}\right]+e^{-i2\pi f (2T_o-t)}+e^{-i2\pi f (2T_o+t)}\right)e^{-u2\pi fT_o} \; df\; du \\
\\
&amp;= -\dfrac{AT_o}{2}\int_{0}^{\infty} \int_{0}^{\infty} u\left(e^{-f(2\pi uT_o - i2\pi  [2T_o+t])}+e^{-f(2\pi uT_o - i2\pi  [2T_o-t])}-2\left[e^{-f(2\pi uT_o -i2\pi  t)}+e^{-f(2\pi u T_o+i2\pi  t)}\right]+e^{-f(2\pi uT_o + i2\pi  [2T_o+t])}+e^{-f(2\pi uT_o + i2\pi  [2T_o-t])}\right) \; df\; du \\
\\
&amp;= -\dfrac{AT_o}{2}\int_{0}^{\infty}  u\left(\dfrac{1}{2\pi uT_o - i2\pi  [2T_o+t]}+\dfrac{1}{2\pi uT_o - i2\pi  [2T_o-t]}-\dfrac{2}{2\pi uT_o -i2\pi  t}-\dfrac{2}{2\pi u T_o+i2\pi  t}+\dfrac{1}{2\pi uT_o + i2\pi  [2T_o+t]}+\dfrac{1}{2\pi uT_o + i2\pi  [2T_o-t]}\right) \; du \\
\\
&amp;= -\dfrac{A}{4\pi}\int_{0}^{\infty}  \dfrac{u}{u - i\left[2+\frac{t}{T_o}\right]}+\dfrac{u}{u - i\left[2-\frac{t}{T_o}\right]}-\dfrac{2u}{u -i\frac{t}{T_o}}-\dfrac{2u}{u+i\frac{t}{T_o}}+\dfrac{u}{u+ i\left[2+\frac{t}{T_o}\right]}+\dfrac{u}{u + i\left[2-\frac{t}{T_o}\right]} \; du \\
&amp;= -\dfrac{A}{2\pi}\int_{0}^{\infty}  \dfrac{u^2}{u^2 +\left[2+\frac{t}{T_o}\right]^2}+\dfrac{u^2}{u^2 + \left[2-\frac{t}{T_o}\right]^2}-\dfrac{2u^2}{u^2 +\left[\frac{t}{T_o}\right]^2}\; du \\
\\
&amp;= -\dfrac{A}{2\pi}\int_{0}^{\infty}  \dfrac{u^2}{u^2 +\left[\frac{t}{T_o}+2\right]^2}+\dfrac{u^2}{u^2 + \left[\frac{t}{T_o}-2\right]^2}-\dfrac{2u^2}{u^2 +\left[\frac{t}{T_o}\right]^2}\; du \\
\\
&amp;= -\dfrac{A}{2\pi} \left(u - \left[\frac{t}{T_o}+2\right] \tan^{-1}\left[\dfrac{u}{\frac{t}{T_o}+2}\right]+u - \left[\frac{t}{T_o}-2\right]\tan^{-1}\left[\dfrac{u}{\frac{t}{T_o}-2}\right]-2u +2\left[\frac{t}{T_o}\right]\tan^{-1}\left[\dfrac{u}{\frac{t}{T_o}}\right]\right)\bigg{|}_0^\infty\\
\\
&amp;= \dfrac{A}{2\pi} \left(\left[\frac{t}{T_o}+2\right] \tan^{-1}\left[\dfrac{u}{\frac{t}{T_o}+2}\right]+\left[\frac{t}{T_o}-2\right]\tan^{-1}\left[\dfrac{u}{\frac{t}{T_o}-2}\right]- 2\left[\frac{t}{T_o}\right]\tan^{-1}\left[\dfrac{u}{\frac{t}{T_o}}\right]\right)\bigg{|}_0^\infty\\
\\
&amp;= \begin{cases} \dfrac{A}{2\pi} \left(-\dfrac{\pi}{2}\left[\dfrac{t}{T_o}+2\right]-\dfrac{\pi}{2}\left[\dfrac{t}{T_o}-2\right]+2\dfrac{\pi}{2}\left[\dfrac{t}{T_o}\right]\right) &amp; \dfrac{t}{T_o} \le -2 \\
\\
\dfrac{A}{2\pi} \left(\dfrac{\pi}{2}\left[\dfrac{t}{T_o}+2\right]-\dfrac{\pi}{2}\left[\dfrac{t}{T_o}-2\right]+2\dfrac{\pi}{2}\left[\dfrac{t}{T_o}\right]\right) &amp;-2 \le \dfrac{t}{T_o}\le 0  \\
\\
\dfrac{A}{2\pi} \left(\dfrac{\pi}{2}\left[\dfrac{t}{T_o}+2\right]-\dfrac{\pi}{2}\left[\dfrac{t}{T_o}-2\right]-2\dfrac{\pi}{2}\left[\dfrac{t}{T_o}\right]\right)&amp;0 \le\dfrac{t}{T_o}\le 2 \\
\\
\dfrac{A}{2\pi} \left(\dfrac{\pi}{2}\left[\dfrac{t}{T_o}+2\right]+\dfrac{\pi}{2}\left[\dfrac{t}{T_o}-2\right]-2\dfrac{\pi}{2}\left[\dfrac{t}{T_o}\right]\right)&amp;\dfrac{t}{T_o}\ge 2\\
\end{cases} \\
\\
h(t) &amp;= \begin{cases} 0 &amp; \dfrac{t}{T_o} \le -2 \\
\\
A\left(1+\dfrac{1}{2}\cdot\dfrac{t}{T_o}\right) &amp;-2 \le \dfrac{t}{T_o}\le 0  \\
\\
A\left(1-\dfrac{1}{2}\cdot\dfrac{t}{T_o}\right)&amp;0 \le\dfrac{t}{T_o}\le 2 \\
\\
0 &amp;\dfrac{t}{T_o}\ge 2\\
\end{cases}\\ 
\\
\end{align*}$$