I'm completely stuck on this one. I only know that it converges thanks to Wolfram, but I don't know how to evaluate it.
$$\int_1^{\infty}\frac{33e^{-\sqrt{x}}}{\sqrt{x}}$$
Thank you for the help.
2026-05-05 23:22:50.1778023370
Solve the improper integral: $\int_1^{\infty}\frac{33e^{-\sqrt{x}}}{\sqrt{x}}$
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Hint. If you make the change of variable $$ u=\sqrt{x},\quad du=\frac{dx}{2\sqrt{x}}, $$ then $$ \int_1^{\infty}\frac{33e^{-\sqrt{x}}}{\sqrt{x}}\:dx=66\int_1^{\infty}e^{-u}\:du. $$