の証明を書きます。
よって、
$\displaystyle\int_0^\pi e^{t\cos x}\cos\left(\frac{x}{2}+t\sin x\right)dx
\\\displaystyle=\frac{1}{2}\int_{-\pi}^\pi\exp\left(\frac{ix}{2}+te^{ix}\right)dx
\\\displaystyle=\frac{1}{2}\int_{-\pi}^\pi e^{\frac{ix}{2}}\sum_{n=0}^\infty\frac{(te^{ix})^n}{n!}dx
\\\displaystyle=\frac{1}{2}\sum_{n=0}^\infty\frac{t^n}{n!}\int_{-\pi}^\pi e^{\left(n+\frac{1}{2}\right)ix}dx
\\\displaystyle=\frac{1}{2}\sum_{n=0}^\infty\frac{t^n}{n!}\frac{e^{\left(n+\frac{1}{2}\right)i\pi}-e^{-\left(n+\frac{1}{2}\right)i\pi}}{i\left(n+\frac{1}{2}\right)}
\\\displaystyle=\sum_{n=0}^\infty\frac{t^n}{n!}\frac{\sin\left(n+\frac{1}{2}\right)\pi}{n+\frac{1}{2}}
\\\displaystyle=\sum_{n=0}^\infty\frac{(-t)^n}{n!}\int_0^1x^{n-\frac{1}{2}}dx
\\\displaystyle=\int_0^1\frac{1}{\sqrt{x}}\sum_{n=0}^\infty\frac{(-tx)^n}{n!}
\\\displaystyle=\int_0^1\frac{e^{-tx}}{\sqrt{x}}
\\\displaystyle=2\int_0^1e^{-tx^2}dx
\\\displaystyle=\int_{-1}^1e^{-tx^2}dx$