I need to determine Fourier Transform of $\delta(2\pi f) = \frac{1}{2 \pi} \delta (f)$ [Note: $\delta$ is Unit impulse function]
Answer is written in my book as follows :
\begin{align} F^{-1}[\delta(2\pi f)] & = \int_{-\infty}^\infty \delta(2\pi f)e^{j2\pi ft} \, df = \frac{1}{2\pi}\int_{-\infty}^\infty \delta(2\pi f)e^{j2\pi ft} \, d(2\pi f) \\[10pt] & =\frac{1}{2\pi} \end{align}
Therefore,
$$\frac{1}{2\pi}\Longleftrightarrow \delta(2 \pi f)$$
$$1 \Longleftrightarrow \delta(f)$$
Here, the left part is in time domain and the right part is in frequency domain. I understood inverse fourier integral part. But couldn't understand how we could deduct the last line from it's upper one. I know that $\int_{-\infty}^\infty \delta (f) e^{j2\pi ft} \, df =1$. I just need to know if there is any way of deducting further Fourier transform pair from an existing one by multiplication or division as it seems to be happened here.
I have found an property that helps in this case.
It is the Linearity property of Fourier Transform.
It says, if
\begin{align} g_1(t)\Longleftrightarrow G_1(f) ~ ~~~~~and ~~~~~~ g_2(t)\Longleftrightarrow G_2(f) \end{align} then, \begin{align} \sum_{k}^{}a_kg_k(t) \Longleftrightarrow \sum_{k}^{}a_kG_k(f) \end{align} For any constants $\{a_k\}$ and signal $\{g_k(t)\}$.
If we take the signal $g_1(t)$ n times then corresponding fourier tansform $G_1(f)$ must be taken n times too.
\begin{align} ng_1(t) \Longleftrightarrow nG_1(f) \end{align} Correspondingly if, $$\frac{1}{2\pi}\Longleftrightarrow \delta(2 \pi f)$$ then using linearity property, (The following line was skipped in the book) $$1 \Longleftrightarrow 2\pi \delta(2 \pi f)$$ But according to question, $\delta(2\pi f) = \frac{1}{2 \pi} \delta (f)$ $$1 \Longleftrightarrow \delta(f) $$