# Using a Simple Integral to Solve a Much Harder One

\begin{align*}
\displaystyle I & = \int_{\alpha}^{\beta} (e)^{x^n} \: dx =
\sum \limits_{\gamma=0}^\infty \left[ \dfrac{(\beta)^{n \gamma +1}-(\alpha)^{n \gamma +1}}{(n \gamma +1) (\gamma ) !} \right]
\\& \\&\therefore IBP \\&
\oplus \Longrightarrow (e)^{x^n} \: ; \: 1 \\&
\ominus \Longrightarrow n(x)^{n-1}(e)^{x^n} \: ; \: x \\& \\&\displaystyle = \left[ x(e)^{x^n} \right] \mid_{\alpha}^{\beta} - \: n\int_{\alpha}^{\beta} (x)^{n}(e)^{x^n} \: dx \\& \\&Note: \left[ x(e)^{x^n} \right] \mid_{\alpha}^{\beta} =
\beta(e)^{\beta^n} - \alpha(e)^{\alpha^n} \Longrightarrow \Delta_1 \\& \\& =\Delta_1 - \: n\int_{\alpha}^{\beta} (x)^{n}(e)^{x^n} \: dx \\& \\&\therefore IBP \\&
\oplus \Longrightarrow (e)^{x^n} \: ; \: (x)^{n} \\&
\ominus \Longrightarrow n(x)^{n-1}(e)^{x^n} \: ; \: \dfrac{(x)^{n+1}}{n+1} \\& \\& =\Delta_1 - \: \left( \dfrac{n}{n+1} \left[ (e)^{x^n}(x)^{n+1} \right] \right) \mid_{\alpha}^{\beta}
+ \: \dfrac{n^2}{n+1} \int_{\alpha}^{\beta} (x)^{2n}(e)^{x^n} \: dx \\& \\&Note: \left( \dfrac{n}{n+1} \left[ (e)^{x^n}(x)^{n+1} \right] \right) \mid_{\alpha}^{\beta} \: \Longrightarrow \Delta_2 \\& \\& \\&\therefore \Delta_k \: Pattern: \:\dfrac{(-n)^{k-1}(x)^{(k-1)n+1}(e)^{x^n}}
{\coprod \limits_{\phi=1}^{k} \left[ (\phi -1) n+1 \right]} \\& \\&\therefore I - \Delta_k \: Pattern: \: \dfrac{(-n)^{k}}
{\coprod \limits_{\phi=1}^{k} \left[ (\phi -1) n+1 \right]}
\int_{\alpha}^{\beta} (x)^{kn}(e)^{x^n} \: dx \\& \\&\therefore I - \Delta_k =\left( \sum \limits_{\gamma=0}^\infty \left[ \dfrac{(x)^{n \gamma +1}}{(n \gamma +1) (\gamma ) !} \right] - \sum \limits_{k=1}^{k} \left[ \dfrac{(-n)^{k-1}(x)^{(k-1)n+1}(e)^{x^n}}
{\coprod \limits_{\phi=1}^{k} \left[ (\phi -1) n+1 \right]} \right] \right) \mid_{\alpha}^{\beta} \\& \\& \\& \\&\therefore \int_{\alpha}^{\beta} (x)^{kn}(e)^{x^n} \: dx =\left(
\dfrac{\coprod \limits_{\phi=1}^{k} \left[ (\phi -1) n+1 \right]}{(-n)^{k}}
\sum \limits_{\gamma=0}^\infty \left[ \dfrac{(x)^{n \gamma +1}}{(n \gamma +1) (\gamma ) !} \right] +
\sum \limits_{k=1}^{k} \left[ \dfrac{(x)^{(k-1)n+1}(e)^{x^n}}{n} \right] \right) \mid_{\alpha}^{\beta} \\& \\&Note: k \geq 1 \: ; \: k \in \mathbb{N}\end{align*}