2 improprer integrals:converge or diverge

93 Views Asked by At

Could you help me understand the correct way to see the behavior of these 2 integrals?

$$\int_{0}^{π/2} \frac {dx}{\tan(x)} $$

In 0: 1/tan(x) ~ 1/x (diverge)

In π/2: is it right to use a direct substitution? Like 1/tan(π/2)=number (converge)

$$\int_{0}^{1} \frac {dx}{ \log{(x-x^2)} }$$

In 0: ~ 1/logx

In1: I don't know

1

There are 1 best solutions below

10
On BEST ANSWER

Note that

$$\int_{0}^{π/2} \frac {dx}{\tan(x)}=\int_{0}^{π/4} \frac {dx}{\tan(x)}+\int_{π/4}^{π/2} \cot x\,dx$$

and the first integral diverges for limit comparison with $\frac1x$.

Note that the second part converges but it is not influent on the result. For the divergence it suffices to note that the first diverges.

For the second note that for $x\to 0^+$ and for $x\to 1^-$

$$\frac {1}{ \log{(x-x^2)}}\to 0$$

thus the integral converges (since the integrand is continuos and bounded on the interval).