English: So again, if we go by the definition, we have the Fourier transform of f(x−a) is equal to $\frac{1}{}\int_{- \infty}^{\infty}\mspace{2mu} f(x - a)e^{i\alpha x}dx$ $\frac{1}{}\int_{- \infty}^{\infty}\mspace{2mu} f(x - a)e^{i\alpha x}dx$కి So, if we substitute here x − a = t that means dx is just simply dt and there will be no change in the limit, so we are getting $\frac{1}{}\int_{- \infty}^{\infty}\mspace{2mu} f(t)e^{i\alpha(t + a)}dt$ and this is exactly what we have the $f\hat{}(\alpha)$ if we take this e^(iαa) out of this integral, then we have $\frac{1}{}\int_{- \infty}^{\infty}\mspace{2mu} f(t)e^{i\alpha t}dt$ which is just the Fourier transform of f(x) $\frac{1}{}\int_{- \infty}^{\infty}\mspace{2mu} f(t)e^{i\alpha(t + a)}dt$ $\frac{1}{}\int_{- \infty}^{\infty}\mspace{2mu} f(t)e^{i\alpha t}dt$ So, here again we have this result that the Fourier transform of this f(x−a) is just e^(iαa) and the Fourier transform of f(x) There is one more property which is called the duality property, which says that the Fourier transform, if we take the Fourier transform of $f\hat{}$, so $f\hat{}$ is the Fourier transform of f(x), so if we take again the Fourier transform of this $f\hat{}$, then we will get back to f but with this  − α, so we can just check again with the help of this inverse Fourier transform, so we know that f(x) is $\frac{1}{}\int_{- \infty}^{\infty}\mspace{2mu} f\hat{}(\alpha)e^{- i\alpha x}d\alpha$ $\frac{1}{}\int_{- \infty}^{\infty}\mspace{2mu} f\hat{}(\alpha)e^{- i\alpha x}d\alpha$ And then we just need to rename that x here, we will rename x to α and the α to x, we will just change their names, so we have here $f(\alpha) = \frac{1}{}\int_{- \infty}^{\infty}\mspace{2mu} f\hat{}(x)e^{- i\alpha x}dx.$ And now we just replace this α by  − α, so we have now this definition of the Fourier transform because in the Fourier transform, we are taking the positive sign there iαx we have $f\hat{}(x)$, so this is the Fourier transform of this $f\hat{}(x)$ $f(\alpha) = \frac{1}{}\int_{- \infty}^{\infty}\mspace{2mu} f\hat{}(x)e^{- i\alpha x}dx$ So, the result is interesting which is called the dual property.
Telugu: కాబట్టి, మనం ఇక్కడ x − a = t ని ప్రతిక్షేపిస్తాము, దీనికి అర్ధం ఏమిటి అంటే dx కేవలం dt మరియు లిమిట్ లలో ఎటువంటి మార్పు ఉండదు, కాబట్టి మనము పొందుతున్నాము మరియు ఇది ఖచ్చితంగా మనకు కావాల్సిన $f\hat{}(\alpha)$ఈ ఇంటిగ్రల్ నుండి e^(iαa)ను బయటకు తీస్తే, మనకు ఉంటుంది, ఇది f (x) యొక్క ఫోరియర్ ట్రాన్స్ఫార్మ్ మాత్రమే.