積分解説01

の証明を書きます。

よって、
$$\begin{gathered} {\displaystyle {\int }_0^{\pi }{e}^{t\cos x}\cos (\frac{x}{2}+t\sin x)dx} \\ {\displaystyle =\frac{1}{2}{\int }_{-\pi }^{\pi }\exp (\frac{ix}{2}+t{e}^{ix})dx} \\ {\displaystyle =\frac{1}{2}{\int }_{-\pi }^{\pi }{e}^{\frac{ix}{2}}\sum _{n=0}^{\mathrm{\infty }}\frac{(t{e}^{ix}{)}^n}{n!}dx} \\ {\displaystyle =\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{t^n}{n!}{\int }_{-\pi }^{\pi }{e}^{(n+\frac{1}{2})ix}dx} \\ {\displaystyle =\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{t^n}{n!}\frac{e^{(n+\frac{1}{2})i\pi }-e^{-(n+\frac{1}{2})i\pi }}{i(n+\frac{1}{2})}} \\ {\displaystyle =\sum _{n=0}^{\mathrm{\infty }}\frac{t^n}{n!}\frac{\sin (n+\frac{1}{2})\pi }{n+\frac{1}{2}}} \\ {\displaystyle =\sum _{n=0}^{\mathrm{\infty }}\frac{(-t{)}^n}{n!}{\int }_0^1{x}^{n-\frac{1}{2}}dx} \\ {\displaystyle ={\int }_0^1\frac{1}{\sqrt{x}}\sum _{n=0}^{\mathrm{\infty }}\frac{(-tx{)}^n}{n!}} \\ {\displaystyle ={\int }_0^1\frac{e^{-tx}}{\sqrt{x}}} \\ {\displaystyle =2{\int }_0^1{e}^{-t{x}^2}dx} \\ {\displaystyle ={\int }_{-1}^1{e}^{-t{x}^2}dx} \end{gathered}$$