I know some properties of Dirac-delta function specially those which are useful for integration involving these functions such as $$\int\limits_{-\infty}^{\infty}\delta(x)\,\mathrm{d}x$$
But I am stuck in this (below) integral and don't know how to proceed further or which property should we use?
$$\int\limits_{-\infty}^\infty e ^{\delta(x)}\,\mathrm{d}x$$
How to solve this problem?
Any hint would be appreciated!
The expression is not defined. However, we can certainly say that the integral does not converge if we define $\delta$ through approximations.
We will use a Lorentz-curve $$\delta_\varepsilon(x) = \frac{1}{\pi}{\frac{\varepsilon}{x^2+\varepsilon^2}}.$$ This function has a narrow maximum at $x=0$, the width is $\sqrt {\epsilon }\to 0$ and the hight $\frac{1}{\sqrt {\epsilon }}\to \infty$. For all $\varepsilon$ the area below the function is $1$. As we choose $\varepsilon\to0$, $\delta_\varepsilon(x)$ tends to $\delta(x)$. Now we can evaluate $$\int\limits_{-\infty}^{\infty}e^{\frac{1}{\pi}\lim\limits_{\varepsilon\to0}{\frac{\varepsilon}{x^2+\varepsilon^2}}}\,\mathrm{d}x > \operatorname{PV} \int\limits_{-\infty}^{\infty}e^{\frac{1}{\pi}\lim\limits_{\varepsilon\to0}{\frac{\varepsilon}{x^2+\varepsilon^2}}}\,\mathrm{d}x,$$ which is not the exact same thing as your integral but it now at least makes a little bit more sense. As you can see I put another integral as an inequality there. The $\operatorname{PV} $ is the Cauchy principal value. So our $x$ does not take values anymore at the "critical point" zero. This integral becomes $$\operatorname{PV} \int\limits_{-\infty}^{\infty}e^{\frac{1}{\pi}\cdot0}\,\mathrm{d}x = \operatorname{PV} \int\limits_{-\infty}^{\infty}1\,\mathrm{d}x\to\infty.$$ This integral is smaller than our original integral and tends to $\infty$, which means that our first integral diverges too.