Gauss積分の一般化

指数法則を使うと，$$\begin{array}{rl}& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\exp \left(\sum _{k=0}^{2m}{c}_k{x}^k\right)dx=e^{c_0}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{c_{2m}{x}^{2m}}\exp \left(\sum _{k=1}^{2m-1}{c}_k{x}^k\right)dx\end{array}$$となり, 指数関数のMaclaurin展開及び，多項定理から, $$\begin{array}{rl}& \exp \left(\sum _{k=1}^{2m-1}{c}_k{x}^k\right)\\ & =\sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{1}{n!}}{\left(\sum _{k=1}^{2m-1}{c}_k{x}^k\right)}^n\\ & =\sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{1}{n!}}\sum _{a_1+\cdots +a_{2m-1}=n}{\displaystyle \frac{n!}{a_1!\cdots a_{2m-1}!}}(c_{1}x{)}^{a_1}\cdots (c_{2m-1}{x}^{2m-1}{)}^{a_{2m-1}}\\ & =\sum _{(a_1,\dots ,a_{2m-1})\in {\mathbb{N}}_0^{2m-1}}{\displaystyle \frac{c_1^{a_1}\cdots c_{2m-1}^{a_{2m-1}}}{a_1!\cdots a_{2m-1}!}}{x}^{a_1+2a_2+\cdots +(2m-1)a_{2m-1}}\\ & =\sum _{(a_1,\dots ,a_{2m-1})\in {\mathbb{N}}_0^{2m-1}}{x}^{a_1+2a_2+\cdots +(2m-1)a_{2m-1}}\prod _{k=1}^{2m-1}{\displaystyle \frac{c_k^{a_k}}{a_k}}\end{array}$$　となります.
これを代入して, 積分と無限和の順序交換を行うと, $$\begin{array}{rl}& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\exp \left(\sum _{k=0}^{2m}{c}_k{x}^k\right)dx\\ & =e^{c_0}\sum _{(a_1,\dots ,a_{2m-1})\in {\mathbb{N}}_0^{2m-1}}\prod _{k=1}^{2m-1}{\displaystyle \frac{c_k^{a_k}}{a_k}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{c_{2m}{x}^{2m}}{x}^{a_1+2a_2+\cdots +(2m-1)a_{2m-1}}dx\\ & =e^{c_0}\sum _{(a_1,\dots ,a_{2m-1})\in {\mathbb{N}}_0^{2m-1}}\left(\prod _{k=1}^{2m-1}{\displaystyle \frac{c_k^{a_k}}{a_k}}\right){\displaystyle \frac{1+(-1{)}^{a_1+2a_2+\cdots +(2m-1)a_{2m-1}}}{2m}}\\ & \times {(-c_{2m})}^{-(1+a_1+2a_2+\cdots +(2m-1)a_{2m-1})/2m}\\ & \times \mathrm{\Gamma }\left({\displaystyle \frac{1+a_1+2a_2+\cdots +(2m-1)a_{2m-1}}{2m}}\right)\end{array}$$ となります. ここで，途中に出てくる$1+(-1{)}^{a_1+2a_2+\cdots +(2m-1)a_{2m-1}}$の部分を見ると，$a_1+2a_2+\cdots +(2m-1)a_{2m-1}$が奇数だと0になるため，これが偶数になる部分だけ取ればよく, $$\begin{array}{rl}& \sum _{k=1}^{2m-1}k{a}_k\equiv 0(\mathrm{mod}.2)\\ ⟺& \sum _{k^{\prime }=1}^m(2k^{\prime }-1)a_{2k^{\prime }-1}+\sum _{k^{″}=1}^{m-1}2k^{″}{a}_{2k^{″}}\equiv 0(\mathrm{mod}.2)\\ ⟺& \sum _{k^{\prime }=1}^m{a}_{2k^{\prime }-1}\equiv 0(\mathrm{mod}.2)\end{array}$$　この条件の下では，$1+(-1{)}^{a_1+2a_2+\cdots +(2m-1)a_{2m-1}}=2$なので, それを適用し，総和・総乗記号を使ってまとめれば, $$\begin{array}{rl}& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\exp \left(\sum _{k=0}^{2m}{c}_k{x}^k\right)dx\\ =& {\displaystyle \frac{e^{c_0}{(-c_{2m})}^{-1/(2m)}}{m}}\\ \times & \sum _{\begin{array}{c}(a_1,\dots ,a_{2m-1})\in {\mathbb{N}}_0^{2m-1}\\ a_1+a_3+\cdots +a_{2m-1}\equiv 0\\ \left(\mathrm{mod}.2\right)\end{array}}\mathrm{\Gamma }({\displaystyle \frac{1}{2m}}+{\displaystyle \frac{1}{2m}}\sum _{k=1}^{2m-1}k{a}_k)\prod _{\ell =1}^{2m-1}{\displaystyle \frac{c_{\ell }^{a_{\ell }}{(-c_{2m})}^{-\ell a_{\ell }/(2m)}}{a_{\ell }!}}\end{array}$$が分かりました.