convergence of integrals

Similarly as one speaks of convergence of series, one can speak of convergence of integrals, especially of Riemann integrals

$${\int }_{I}f(t)𝑑t.$$

This integral is convergent, if it exists, and otherwise divergent.  One can also speak of absolute convergence of integrals.

Example.  Study the convergence of the integral

${\displaystyle {\int }_1^2}{\displaystyle \frac{dx}{{(\ln x)}^c}}$(1)

where $c$ is a real constant.

According to the logarithm series, we may write for  ,  where $b$ is sufficiently close to $1$, the estimations

$\ln (x-1)=x-1+O({(x-1)}^2)=(x-1)[1+O(x-1)]\{\begin{array}{cc}\le 2(x-1),\hfill & \\ \ge \frac{1}{2}(x-1).\hfill & \end{array}$

Let  . 

$1^{\circ }$.  For  $c>1$:

${\displaystyle {\int }_a^b}{\displaystyle \frac{dx}{{(\ln x)}^c}}$$\geqq {\displaystyle {\int }_a^b}{\displaystyle \frac{dx}{2^c{(x-1)}^c}}=-{\displaystyle \frac{1}{2^c}}\underset{x=a}{\overset{b}{/}}{\displaystyle \frac{1}{(c-1){(x-1)}^{c-1}}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{2^c(c-1)}}\left[{\displaystyle \frac{1}{{(a-1)}^{c-1}}}-{\displaystyle \frac{1}{{(b-1)}^{c-1}}}\right]⟶\mathrm{\infty }\mathit{\hspace{1em}}\text{as}\mathit{\hspace{1em}}a\to 1+$

$2^{\circ }$.  For  $c=1$:

${\displaystyle {\int }_a^b}{\displaystyle \frac{dx}{\ln x}}$$\geqq {\displaystyle {\int }_a^b}{\displaystyle \frac{dx}{2(x-1)}}={\displaystyle \frac{1}{2}}\underset{a}{\overset{b}{/}}\ln (x-1)$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{2}}\left[\ln (b-1)-\ln (a-1)\right]⟶\mathrm{\infty }\mathit{\hspace{1em}}\text{as}\mathit{\hspace{1em}}a\to 1+$

$3^{\circ }$.  For  :

$\leqq {\displaystyle {\int }_a^b}{\displaystyle \frac{2^{c}dx}{{(x-1)}^c}}={\mathrm{\hspace{0.33em}2}}^c\underset{x=a}{\overset{b}{/}}{\displaystyle \frac{x^{1-c}}{1-c}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{2^c}{1-c}}\left[{(b-1)}^{1-c}-{(a-1)}^{1-c}\right]⟶{\displaystyle \frac{2^c}{1-c}}{(b-1)}^{1-c}\mathit{\hspace{1em}}\text{as}\mathit{\hspace{1em}}a\to 1+$

Consequently, the integral ${\int }_a^b\frac{dx}{{(\ln x)}^c}$, and thus also (1), converges if and only if  .