Dirac delta integral with $\delta(\infty) \cdot e^{\infty}$

Solution 1:

PRIMER ON THE DIRAC DELTA AS A GENERALIZED FUNCTION

The Dirac Delta and the Unit Doublet (the so-called "derivative" of the Dirac Delta) are not functions. Rather, they are Generalized Functions, also known as Distributions.

Distributions are linear Functionals that map test functions (smooth functions) into numbers, whereas a function maps numbers into numbers. For the Dirac Delta, the functional definition is given as

$$\langle f,\delta_a\rangle =f(a) \tag 1$$

where $f$ is a suitable test function.

Now, in practice, we often write the functional notation in $(1)$ formally as

$$\langle f,\delta_a\rangle=\int_{-\infty}^{\infty}f(x)\delta(x-a)\,dx \tag 2$$

But the object on the right-hand side of $(2)$ is actually not an integral. And the evaluation of $\delta (x-a)$ as a function is meaningless. In practice, we often see the Dirac Delta defined at points by

$$\delta(x)=\begin{cases}0&,x\ne 0\\\\\infty&,x=0\end{cases}$$

but this is obvious nonsense. Rather, the interpretation here can be made physical through a regularization of the Dirac Delta wherein there is a family of functions $\delta_n(x)\in C^\infty_C$ for which

$$\lim_{n\to \infty}\delta_n(x)= \begin{cases} 0&, x\ne 0\\\\ \infty&,x=0\end{cases}$$

and

$$\lim_{n\to \infty}\int_{-\infty}^\infty f(x)\delta_n(x-a)\,dx=f(a)$$

for all suitable test functions $f$. We can formally write this regularization of the delta function as

$$\delta(x)\sim\lim_{n\to \infty}\delta_n(x)$$

Therefore, we interpret the integral notation for the functional relationship in $(2)$ as

$$\int_{-\infty}^{\infty} f(x) \delta(x-a)\,dx=\lim_{n\to \infty}\int_{-\infty}^\infty f(x)\delta_n(x-a)\,dx$$

For more on the Dirac Delta, see THIS ANSWER, THIS ONE, THIS ONE, and THIS ONE.

THE UNIT DOUBLET AS A GENERALIZED FUNCTION

The Unit Doublet $\delta'$ is defined as in terms of the Dirac Delta as

$$\langle f,\delta_a'\rangle=-\langle f',\delta_a\rangle=-f'(a)$$

It is, therefore, a functional that maps a test function $f$ into $-f'$. We can formally, write

$$\langle f,\delta_a'\rangle=\int_{-\infty}^{\infty}f(x)\delta'(x-a)\,dx=-\int_{-\infty}^{\infty}f'(x)\delta(x-a)\,dx=-f'(a)$$

and proceed heuristically as with the Dirac Delta. Note that if $f(x)=e^{x^2}$ and $a=3$, we have immediately that

$$\langle e^{x^2},\delta'_3\rangle =-\left.\frac{de^{x^2}}{dx}\right|_{x=3}=-6e^9$$

Solution 2:

Using the shifting property $f(x) \, \delta(x-a) = f(a) \, \delta(x-a)$ then \begin{align} \int \delta^{\prime}(x-a) \, f(x) \, dx &= \left[ f(x) \, \delta(x-a) \right] - \int \delta(x-a) \, f^{\prime}(x) \, dx \\ &= \left[ f(a) \, \delta(x-a) \right] - f^{\prime}(a) \end{align} For the case of $f(x) = e^{x^2}$ then $f^{\prime}(x) = 2 \, x \, e^{x^2}$. For the limits $(-\infty, \infty)$ the $\delta$ function is zero at the end points being evaluated. Now, \begin{align} \int_{-\infty}^{\infty} \delta^{\prime}(x-a) \, e^{x^{2}} \, dx &= \left[ e^{a^{2}} \, \delta(x-a) \right]_{-\infty}^{\infty} - 2 \, a \, e^{a^{2}} \\ &= - 2 \, a \, e^{a^{2}}. \end{align}

Shifting property: $$ \int \delta(x-a) \, f(x) \, dx = f(a) = \int f(a) \, \delta(x-a) \, dx $$ or $$ \int \left[ f(x) \, \delta(x-a) - f(a) \, \delta(x-a) \right] \, dx = 0.$$ In order for the general integral to have a zero result then the integrand must be zero which leads to $$f(x) \, \delta(x-a) = f(a) \, \delta(x-a)$$