Question

\(5 - 32\)Determine whether each integral is convergent or divergent. Evaluate those that are convergent.

15.\(\int_{ - \infty }^0 z {e^{2z}}dz\)

Short answer

Integral is convergent and its value is\(\frac{{ - 1}}{4}\).

Step-by-step solution

Definition for converging integral

Consider the integral\(\int_{\bf{a}}^{\bf{t}} {\bf{f}} {\bf{(x)dx}}\)is for the limit with the finite number\(t \ge a\). Then, the integral is written as:

\(\int_a^\infty f (x)dx = \mathop {\lim }\limits_{t \to \infty } \int_a^t f (x)dx\)

Evaluate the integral.

Write the given integral using the definition mentioned above.

\(\int_{ - \infty }^0 z {e^{2z}}dz = \mathop {\lim }\limits_{t \to - \infty } \int_t^0 z {e^{2z}}dz.\)

Use the integration by parts i.e; \(\int u dv = uv - \int v du{\rm{.}}\)

Here\(u = z,dv = {e^{2z}} \to du = dz,dv = \frac{1}{2}{e^{2z}}{\rm{ }}\)

\(\begin{aligned}{c}\int z {e^{2z}} &= z\left( {\frac{{{e^{2z}}}}{2}} \right) - \frac{1}{2}\int {{e^{2z}}} dz\\ &= z\left( {\frac{{{e^{2z}}}}{2}} \right) - \frac{{{e^{2z}}}}{4}\end{aligned}\)

Evaluate at\(\infty \):

Now apply the limits.

\(\begin{aligned}{c}\mathop {\lim }\limits_{t \to - \infty } \int_t^0 z {e^{2z}}dz &= \mathop {\lim }\limits_{t \to - \infty } \left( {z\left( {\frac{{{e^{2z}}}}{2}} \right) - \frac{{{e^{2z}}}}{4}} \right)_t^0\\ &= \mathop {\lim }\limits_{t \to - \infty } \left( { - \frac{1}{4} - \frac{{t{e^{2t}}}}{2} + \frac{{{e^{2t}}}}{4}} \right)\\ &= - \frac{1}{4}\end{aligned}\)

The value of integral is finite so it is convergent.