English: $\frac{1}{}\int_{- \infty}^{\infty}\mspace{2mu} e^{- kt\alpha^{2} - ixa}d\alpha$ So, our g(x) will be $\frac{1}{}$ and from here we have and $e^{- \frac{x^{2}}{4kt}.\ }\ \ \ \ \ \ \ So,\ in\ this\ way,\ we\ got\ our\ \ $ Just to summarize again, what we have done our u(x,t) was $\frac{1}{}\int_{- \infty}^{\infty}\mspace{2mu} f\hat{}(\alpha)e^{- k\alpha^{2}t}e^{- i\alpha x}d\alpha$.
Telugu: $\frac{1}{}\int_{- \infty}{\infty}\mspace{2mu} e^{- kt\alpha^{2} - ixa}d\alpha$ కాబట్ట int_{1}ి, మా g(x) $\frac{1}{}$ మరియు ఇక్కడ నుండి మాకు $e^{{- \frac{x^{2}}{4kt}.\ \{2} in\ \\ \in\ \ \