簡単な無限級数の積分表示

$${\int }_0^1{\displaystyle \frac{x^{a+b-1}(-x^a{)}^n}{x^a+1}}dx=(-1{)}^n{\int }_0^1{\displaystyle \frac{x^{a+b-1}(x^a{)}^n}{x^a+1}}dx$$
$a+b-1$は条件より,必ず正である.よって,積分区間で$0\leqq x^{a+b-1}\leqq 1$全てに$x=0$を代入して左辺,単調増加より右辺ができる.
$$0<{\int }_0^1{\displaystyle \frac{x^{a+b-1}(x^a{)}^n}{x^a+1}}dx<{\int }_0^1{\displaystyle \frac{(x^a{)}^n}{x^a+1}}dx$$
$${\int }_0^1{\displaystyle \frac{(x^a{)}^n}{x^a+1}}dx<{\int }_0^1{x}^{an}dx={\displaystyle \frac{1}{an+1}}$$

$$\sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{(-1{)}^n}{an+b}}=\underset{n\to \mathrm{\infty }}{lim}\sum _{k=0}^n{\displaystyle \frac{(-1{)}^k}{ak+b}}$$
とする.
$$\sum _{k=0}^n{\displaystyle \frac{(-1{)}^k}{ak+b}}={\displaystyle \frac{1}{b}}-{\displaystyle \frac{1}{a+b}}+{\displaystyle \frac{1}{2a+b}}-{\displaystyle \frac{1}{3a+b}}+\cdots +(-1{)}^n{\displaystyle \frac{1}{an+b}}$$
等式$(m\ne -1)$
$${\displaystyle \frac{1}{m+1}}={\int }_0^1{x}^{m}dx$$
より,
$$\sum _{k=0}^n{\displaystyle \frac{(-1{)}^k}{ak+b}}={\int }_0^1{x}^{b-1}-x^{a+b-1}+x^{2a+b-1}-x^{3a+b-1}+\cdots +(-1{)}^n{x}^{an+b-1}dx={\int }_0^1\sum _{k=0}^n{x}^{b-1}(-x^a{)}^{k}dx$$
右辺の被積分関数は初項が$x^{b-1}$公比が$-x^a$の等比数列の初項から第$n$項までの和である.
$${\int }_0^1\sum _{k=0}^n{x}^{b-1}(-x^a{)}^{k}dx={\int }_0^1{\displaystyle \frac{x^{b-1}+x^{a+b-1}(-x^a{)}^n}{x^a+1}}dx={\int }_0^1{\displaystyle \frac{x^{b-1}}{x^a+1}}dx+{\int }_0^1{\displaystyle \frac{x^{a+b-1}(-x^a{)}^n}{x^a+1}}dx$$
補題1より,$n\to \mathrm{\infty }$すると,
$$\sum _{k=0}^{\mathrm{\infty }}{\displaystyle \frac{(-1{)}^n}{an+b}}={\int }_0^1{\displaystyle \frac{x^{b-1}}{x^a+1}}dx$$
が成り立つ.