I'm trying to show the convergence or divergence of the following integral:
$$ \int_{0}^1 \frac{\sec^2(x)}{x\sqrt{x}} dx$$
I begin by writing it as:
$$ \int_{0}^1 \frac{1+\tan^2(x)}{x^{3/2}} dx$$
And this is the step I'm not entirely sure is correct:
$$ \int_{0}^1\frac{1}{x^{3/2}}dx +\int_{0}^1\frac{\tan^2(x)}{x^{3/2}}dx$$
Assuming it is correct, the integral would be:
$\lim_{t\to 0^{+}}\int_{t}^1\frac{1}{x^{3/2}}dx +\int_{t}^1\frac{\tan^2(x)}{x^{3/2}}dx$
And by just evaluating $\lim_{t\to 0^{+}}\int_{t}^1\frac{1}{x^{3/2}}dx$ the divergence could be shown.
$\lim_{t\to 0^{+}}\int_{t}^1\frac{1}{x^{3/2}}dx = \lim_{t\to 0^{+}}\Bigg[\frac{-2}{\sqrt{x}} \Bigg]_{t}^{1}$
$=\lim_{t\to 0^{+}}\frac{-2}{\sqrt{1}} - (\frac{-2}{\sqrt{t}}) $
$=\lim_{t\to 0^{+}}-2 + \frac{2}{\sqrt{t}} $
This limit is $\infty$ so the integral does not converge.
Is this correct? Can I split the integral like it is done with a proper definite integral?
Since the improper integral is a limit, you can split it into a sum, provided both summands have a limit: just apply $$ \lim_{t\to0}\int_t^1 (f(x)+g(x))\,dx= \lim_{t\to0}\int_t^1 f(x)\,dx+ \lim_{t\to0}\int_t^1 g(x)\,dx $$ which you can do provided both limits on the right-hand side exist, with the usual rules for $\infty$.
If one of the limits is $\infty$ and the other is either finite or $\infty$, then you can conclude for the divergence of the left-hand side. Similarly for both $-\infty$ or one $-\infty$ and the other one finite.
But if one is $\infty$ and the other one is $-\infty$ you can't conclude.