The integral shown in the image is:

$\int_{-\infty}^{-1} \frac{1}{\sqrt[7]{x}} \, dx$

This represents an improper integral because one of the limits is negative infinity. To evaluate this, we need to express the integrand more conveniently and then handle the limits.

1. Rewrite the integrand: $\frac{1}{\sqrt[7]{x}} = x^{-\frac{1}{7}}$

Thus, the integral becomes:

$\int_{-\infty}^{-1} x^{-\frac{1}{7}} \, dx$

2. Evaluate the indefinite integral:

$\int x^{-\frac{1}{7}} \, dx = \frac{x^{\frac{6}{7}}}{\frac{6}{7}} = \frac{7}{6} x^{\frac{6}{7}}$

3. Evaluate the limits: The improper integral requires careful treatment of the infinite lower limit. The integral is:

$\lim_{a \to -\infty} \left[ \frac{7}{6} x^{\frac{6}{7}} \right]_{a}^{-1}$

Now let's check the behavior at both bounds.

• At $x = -1$: $\frac{7}{6}(-1)^{\frac{6}{7}} = \frac{7}{6} \times (-1)^{6/7}$

Here, we encounter an issue because the exponent $\frac{6}{7}$ is a fractional exponent involving a negative base, which is problematic. This makes the integral complex and suggests the integral diverges in the real number system.

Conclusion: The integral does not converge within the real number system because the function $x^{-\frac{1}{7}}$ is not defined for negative values of $x$, causing divergence or complex results.

Would you like more detailed steps or have any questions?

Here are 5 related questions you might find interesting:

1. How do we handle improper integrals with infinity as a bound?
2. What are the conditions under which an integral diverges?
3. How can we evaluate integrals involving negative powers of $x$?
4. What is the difference between convergence and divergence in integrals?
5. Can we convert divergent integrals into convergent forms using complex analysis?

Tip: Always check the domain of the integrand before solving an improper integral to avoid undefined behavior or divergence.