∫ α + ∞ 1 ( x − 2 ) | x − 3 | d x {\displaystyle \int _{\alpha }^{+\infty }{{\frac {1}{(x-2){\sqrt {|x-3|}}}}\,dx}}
∑ 1 + ∞ 1 n ( n 2 − 4 n + 1 ) {\displaystyle \sum _{1}^{+\infty }{\frac {1}{n(n^{2}-4n+1)}}} e = ∫ n + ∞ 1 n ( n 2 − 4 n + 1 ) d x {\displaystyle e=\int _{n}^{+\infty }{{\frac {1}{n(n^{2}-4n+1)}}\,dx}}
a 0 = ∫ − π 0 − x d x + ∫ 0 π x d x {\displaystyle a_{0}=\int _{-\pi }^{0}-xdx+\int _{0}^{\pi }xdx}
