Question

integral_1^∞ 1/(x-3)^(3/2) dx

Answer 1

The integrand is undefined when \(x=3\), so you have to write as two integrals first:
\[\int_1^3\frac{dx}{(x-3)^{3/2}}+\int_3^\infty\frac{dx}{(x-3)^{3/2}}\]
These are improper integrals, so write as limits:
\[\lim_{a\to3^-}\int_1^a\frac{dx}{(x-3)^{3/2}}+\lim_{b\to3^+}\lim_{c\to\infty}\int_b^c\frac{dx}{(x-3)^{3/2}}\]
Apply the power rule to integrate:
\[\lim_{a\to3^-}\left[-\frac{2}{(x-3)^{1/2}}\right]_1^a+\lim_{b\to3^+}\lim_{c\to\infty}\left[-\frac{2}{(x-3)^{1/2}}\right]_b^c\]
\[\left[\left(\lim_{a\to3^-}-\frac{2}{(a-3)^{1/2}}\right)-\left(-\frac{2}{(1-3)^{1/2}}\right)\right]+\cdots\]
This limit is \(\infty\), which means this integral does not converge, which means the original integral does not converge.