Question

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

14. \(\int_{ - \infty }^\infty {{x^2}} {e^{ - {x^3}}}dx\)

Short answer

Integral is divergent.

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:

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

Break the integral

Break the given integral at 0 & write it as the sum of two one sided improper integrals.

\(\int_{ - \infty }^\infty {{x^2}} {e^{ - {x^3}}}dx = \int_{ - \infty }^0 {{x^2}} {e^{ - {x^3}}}dx + \int_0^\infty {{x^2}} {e^{ - {x^3}}}dx\;\;\;\)

Here\(a = 0\).

Now, evaluate the integrals on the right side separately:

Let\(u = {x^3}\)then,

\(\begin{aligned}{l}du &= 3{x^2}dx\\{x^2}dx &= \frac{1}{3}du\end{aligned}\)

Evaluate\(\int_0^\infty  {{x^2}} {e^{ - {x^3}}}dx\)

Using the substitution given above we have:

\(\begin{aligned}{c}\int_0^\infty {{x^2}} {e^{ - {x^3}}}dx &= \mathop {\lim }\limits_{t \to \infty } \int_0^t {\left( {{x^2}{e^{ - {x^3}}}} \right)} dx\\ &= \mathop {\lim }\limits_{t \to \infty } \int_0^t {\left( {\frac{1}{3}{e^{ - u}}} \right)} du\\ &= \mathop {\lim }\limits_{t \to \infty } \left( {\frac{1}{3}\left( { - {e^{ - u}}} \right)_0^t} \right)\\ &= \mathop {\lim }\limits_{t \to \infty } \left( {\frac{1}{3}\left( { - {e^{ - t}} + {e^0}} \right)} \right)\end{aligned}\)

On simplifying we get:

\(\begin{aligned}{c}\int_0^\infty {{x^2}} {e^{ - {x^3}}}dx &= \frac{1}{3}(1 - 0)\;\;\;{\rm{ Since, }}{e^{ - \infty }} &= 0.\\ &= \frac{1}{3}\end{aligned}\)

So, this integral is convergent.

Evaluate\(\int_{ - \infty }^0 {{x^2}} {e^{ - {x^3}}}dx\)

The first integral will become:

\(\begin{aligned}{c}\int_{ - \infty }^0 {{x^2}} {e^{ - {x^3}}}dx &= \mathop {\lim }\limits_{t \to - \infty } \int_t^0 {\left( {{x^2}{e^{ - {x^3}}}} \right)} dx\\ &= \mathop {\lim }\limits_{t \to - \infty } \int_t^0 {\left( {\frac{1}{3}{e^{ - u}}} \right)} dx\\ &= \mathop {\lim }\limits_{t \to - \infty } \left( {\frac{1}{3}\left( { - {e^{ - t}}} \right)_t^0} \right)\\ &= \mathop {\lim }\limits_{t \to - \infty } \left( {\frac{1}{3}\left( { - {e^0} + {e^{ - t}}} \right)} \right)\end{aligned}\)

Solve further as,

\(\begin{aligned}{c}\int_{ - \infty }^0 {{x^2}} {e^{ - {x^3}}}dx &= \mathop {\lim }\limits_{t \to - \infty } \left( {\frac{1}{3}\left( {{e^{ - t}} - 1} \right)} \right)\\ &= \frac{1}{3}\left( {{e^\infty } - 1} \right)\\ &= \infty {\rm{ Since, }}{e^\infty } &= \infty \end{aligned}\)

So, the first improper integral is divergent.

Therefore, improper integral \(\int_{ - \infty }^\infty {{x^2}} {e^{ - {x^3}}}dx\)is divergent.