Improper Riemann integral questions and determining whether they exist.

Despite the imtegrand becoming $\infty$ at say $(x=a)$. One way it becomes improper but convergent is when $0<p<1$ where $$I=\int_{a}^{b} \frac{dx}{(x-a)^p}<\infty.$$ For instance $$\int_{0}^{1} \frac{dx}{\sqrt{x}}=2$$ is improper but convergent. The integral $$\int_{0}^{1} \frac{dx}{x^{1.01}} =\infty,$$ is improper and divergent.

Next, one more way and integral becomes improper but convergent when $q>1$ where the integral is $$J=\int_{1}^{\infty} \frac{dx}{x^q}.$$ For instance $$\int_{1}^{\infty} \frac{dx}{x^{1.01}}=100, \int_{0}^{\infty} \frac{dx}{\sqrt{x}}=\infty$$

Finally both integrals of yours are inproper but divergent the first one is so as $p=1$ and second one is so as $q=1$.

Notice that the well known integral $$\pi=\int{-1}^{1} \frac{dx}{\sqrt{1-x^2}}=\int_{-1}^{1} \frac{dx}{(1-x)^{1/2} (1+x)^{1/2}},$$ is improper but convergent because for both $x=\pm 1$, $p=1/2<1.$

The integral $$\int_{0}^{1} \ln x~ dx =x\ln x-x|_{0}^{1}=-1.$$ is improper but convergent, In this case one has to take the limit $x\rightarrow 0$ of the anti-derivative.

The improper integrals do not connect well to area under the curve. This may also be seen as a defect in the theory of integration or these integrals are called improper which may or may not be convergent.

There are many other ways and integral is improper but converget for instance $$\int_{0}^{\infty} \frac{\sin x}{x} dx=\pi/2,$$ despite the integrand existing only as limit when $x \rightarrow 0.$

The integral $$\int_{0}^{\infty} \frac {dx}{(1+x^4)^{1/4}} \sim \int^{\infty} \frac{dx}{(x^4)^{1/4}} =\infty$$ is divergen, we know this without actually solving this integral!

Most often studying the integrand near the point of singularity (discontinuity) or $\infty$ and making out the value of $p$ or $q$ helps in finding if it is convergent.