Since the Maclaurin series for e y {\displaystyle e^{y}} is
∑ n = 0 ∞ y n n ! , {\displaystyle \sum _{n=0}^{\infty }{\frac {y^{n}}{n!}},}
the Maclaurin series for e − x 2 {\displaystyle e^{-x^{2}}} is
∑ n = 0 ∞ ( − x 2 ) n n ! = ∑ n = 0 ∞ ( − 1 ) n x 2 n n ! = 1 − x 2 + 1 2 x 4 − 1 6 x 6 + … {\displaystyle \sum _{n=0}^{\infty }{\frac {(-x^{2})^{n}}{n!}}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{n!}}=1-x^{2}+{\frac {1}{2}}x^{4}-{\frac {1}{6}}x^{6}+\dots }
Hence, the first few nonzero terms of the Maclaurin series for the integrand e − x 2 − 1 x {\displaystyle {\frac {e^{-x^{2}}-1}{x}}} are
1 x ( ( 1 − x 2 + 1 2 x 4 − 1 6 x 6 + … ) − 1 ) = − x + 1 2 x 3 − 1 6 x 5 + … . {\displaystyle {\begin{aligned}{\frac {1}{x}}\left(\left(1-x^{2}+{\frac {1}{2}}x^{4}-{\frac {1}{6}}x^{6}+\dots \right)-1\right)&=-x+{\frac {1}{2}}x^{3}-{\frac {1}{6}}x^{5}+\dots .\end{aligned}}}
Anti-differentiating these, we obtain the first few nonzero terms of the indefinite integral ∫ e − x 2 − 1 x d x , {\displaystyle {\begin{aligned}\int {\frac {e^{-x^{2}}-1}{x}}\,dx,\end{aligned}}}
C − x 2 2 + 1 8 x 4 − 1 36 x 6 + … {\displaystyle C-{\frac {x^{2}}{2}}+{\frac {1}{8}}x^{4}-{\frac {1}{36}}x^{6}+\dots }
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