To evaluate the integral

$\int_1^{\infty} \frac{e^{\sqrt{x-1}}}{\sqrt{x-1}} \, dx,$

we can use a substitution. Let:

$u = \sqrt{x-1} \implies x = u^2 + 1 \implies dx = 2u \, du.$

The limits change as follows:

• When $x = 1$, $u = 0$.
• When $x \to \infty$, $u \to \infty$.

Now, substituting these into the integral:

$\int_0^{\infty} \frac{e^u}{u} (2u) \, du = 2 \int_0^{\infty} e^u \, du.$

This simplifies to:

$2 \int_0^{\infty} e^u \, du.$

However, the integral

$\int_0^{\infty} e^u \, du$

diverges, because $e^u$ grows exponentially as $u \to \infty$. Thus, the original integral also diverges.

Therefore, the integral

$\int_1^{\infty} \frac{e^{\sqrt{x-1}}}{\sqrt{x-1}} \, dx$

diverges.

Summary

• The integral diverges.

Would you like further details or explanations on any part of this process? Here are some questions for further exploration:

1. How do substitution methods work in integrals?
2. What are the implications of an integral diverging?
3. Can you provide examples of convergent integrals involving exponential functions?
4. What other techniques can be used to evaluate improper integrals?
5. How does the behavior of $e^{\sqrt{x-1}}$ impact convergence?

Tip: Always check the behavior of the integrand as it approaches the limits of integration to determine convergence or divergence.