Question

If \(\int_{ - \infty }^a f (x)dx\)is convergent and a and b are real numbers, Show that\(\int_{ - \infty }^a f (x)dx + \int_a^\infty f (x)dx = \int_{ - \infty }^b f (x)dx + \int_b^\infty f (x)dx\).

Short answer

Proved

Step-by-step solution

Given Data

The given integral\(\int_{ - \infty }^\infty f (x)dx\)is continuous.

The values of \(a\) and \(b\) are real numbers.

Evaluating the Given Expression

Consider the value\(a < b\).

Show the integral as follows:

Calculating RHS of Eq.1

\(\begin{aligned}{c}\int_{ - \infty }^\infty f (x)dx &= \int_{ - \infty }^a f (x)dx + \int_a^\infty f (x)dx.......(1)\\\int_{ - \infty }^a f (x)dx + \int_a^\infty f (x)dx &= \mathop {\lim }\limits_{t \to - \infty } \int_t^a f (x)dx + \mathop {\lim }\limits_{s \to \infty } \int_a^s f (x)dx\\ &= \mathop {\lim }\limits_{t \to - \infty } \int_t^a f (x)dx + \mathop {\lim }\limits_{s \to \infty } \left( {\int_a^b f (x)dx + \int_b^s f (x)dx} \right)\\ &= \mathop {\lim }\limits_{t \to - \infty } \int_t^a f (x)dx + \int_a^b f (x)dx + \mathop {\lim }\limits_{s \to \infty } \left( {\int_b^s f (x)dx} \right)\\ &= \left( {\mathop {\lim }\limits_{t \to - \infty } \int_t^a f (x)dx + \int_a^b f (x)dx} \right) + \int_b^\infty f (x)dx\\\int_{ - \infty }^a f (x)dx + \int_a^\infty f (x)dx &= \left( {\mathop {\lim }\limits_{t \to - \infty } \int_t^a f (x)dx + \mathop {\lim }\limits_{t \to - \infty } \int_a^b f (x)dx} \right) + \int_b^\infty f (x)dx\\ &= \left( {\mathop {\lim }\limits_{t \to - \infty } \int_t^b f (x)dx} \right) + \int_b^\infty f (x)dx\\ &= \int_{ - \infty }^b f (x)dx + \int_b^\infty f (x)dx\end{aligned}\)

Thus, the expression\(\int_{ - \infty }^a f (x)dx + \int_a^\infty f (x)dx = \int_{ - \infty }^b f (x)dx + \int_b^\infty f (x)dx\)is proved.