# Integral$\int_{-\infty}^\infty x^{2n} e^{-\beta (x^2+\cos x+\alpha x)}dx$

Hi I am trying to integrate $$\int_{-\infty}^\infty\int_{-\infty}^\infty (xy)^{2n}\exp\left({-\beta(x^2+y^2+\cos x+\alpha x+iy)}\right)dxdy \quad \alpha,\beta,n >0.$$ These integrals can be broken up into the forms $$\int_{-\infty}^\infty x^{2n} e^{-\beta (x^2+\cos x+\alpha x)}dx\int_{-\infty}^\infty y^{2n} e^{-\beta (y^2+iy)}dy.$$ I can do the integral over y, but I am still stuck on $$I=\int_{-\infty}^\infty x^{2n} e^{-\beta (x^2+\cos x+\alpha x)}dx$$

• Hi, Jeff. I've seen that you've posted many questions over a short period of time. Hence, I wish to tell you that there is a limit to the number of questions that can be posted by a user over a specific period of time. Please take this into consideration for the next question that you post. – user122283 Apr 9 '14 at 23:01
• @SanathDevalapurkar oh yeah whats that limit? – Jeff Faraci Apr 9 '14 at 23:02
• $50$ questions per $30$ days. – user122283 Apr 9 '14 at 23:04
• okay that doesn't apply here, thanks! – Jeff Faraci Apr 9 '14 at 23:04
• $\cosh x$ or $\cos x$ ? I saw you changed from $\cosh x$ to $\cos x$ . (math.stackexchange.com/posts/747451/revisions) – Harry Peter Apr 29 '14 at 16:36

$$\text{If you really want to ask about }\int_{-\infty}^\infty x^{2n}e^{-\beta(x^2+\cos x+\alpha x)}~dx~:$$

Hint:

$$\int_{-\infty}^\infty x^{2n}e^{-\beta(x^2+\cos x+\alpha x)}~dx$$

$$=\int_{-\infty}^\infty x^{2n}e^{-\beta(x^2+\alpha x)}e^{-\beta\cos x}~dx$$

$$=\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\dfrac{\beta^{2p}x^{2n}e^{-\beta(x^2+\alpha x)}\cos^{2p}x}{(2p)!}dx-\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\dfrac{\beta^{2p+1}x^{2n}e^{-\beta(x^2+\alpha x)}\cos^{2p+1}x}{(2p+1)!}dx$$

$$=\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\dfrac{\beta^{2p}x^{2n}e^{-\beta(x^2+\alpha x)}}{4^p(p!)^2}dx+\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\sum_{q=1}^p\dfrac{\beta^{2p}x^{2n}e^{-\beta(x^2+\alpha x)}\cos2qx}{2^{2p-1}(p-q)!(p+q)!}dx-\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\sum_{q=0}^p\dfrac{\beta^{2p+1}x^{2n}e^{-\beta(x^2+\alpha x)}\cos((2q+1)x)}{4^p(p-q)!(p+q+1)!}dx$$ (according to http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formula)

$$=\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\dfrac{\beta^{2p}x^{2n}e^{-\beta\left(x^2+\alpha x+\frac{\alpha^2}{4}-\frac{\alpha^2}{4}\right)}}{4^p(p!)^2}dx+\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\sum_{q=1}^p\dfrac{\beta^{2p}x^{2n}e^{-\beta\left(x^2+\alpha x+\frac{\alpha^2}{4}-\frac{\alpha^2}{4}\right)}\cos2qx}{2^{2p-1}(p-q)!(p+q)!}dx-\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\sum_{q=0}^p\dfrac{\beta^{2p+1}x^{2n}e^{-\beta\left(x^2+\alpha x+\frac{\alpha^2}{4}-\frac{\alpha^2}{4}\right)}\cos((2q+1)x)}{4^p(p-q)!(p+q+1)!}dx$$

$$=\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\dfrac{\beta^{2p}e^\frac{\alpha^2\beta}{4}x^{2n}e^{-\beta\left(x+\frac{\alpha}{2}\right)^2}}{4^p(p!)^2}dx+\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\sum_{q=1}^p\dfrac{\beta^{2p}e^\frac{\alpha^2\beta}{4}x^{2n}e^{-\beta\left(x+\frac{\alpha}{2}\right)^2}\cos2qx}{2^{2p-1}(p-q)!(p+q)!}dx-\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\sum_{q=0}^p\dfrac{\beta^{2p+1}e^\frac{\alpha^2\beta}{4}x^{2n}e^{-\beta\left(x+\frac{\alpha}{2}\right)^2}\cos((2q+1)x)}{4^p(p-q)!(p+q+1)!}dx$$

$$=\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\dfrac{\beta^{2p}e^\frac{\alpha^2\beta}{4}\left(x-\dfrac{\alpha}{2}\right)^{2n}e^{-\beta x^2}}{4^p(p!)^2}dx+\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\sum_{q=1}^p\dfrac{\beta^{2p}e^\frac{\alpha^2\beta}{4}\left(x-\dfrac{\alpha}{2}\right)^{2n}e^{-\beta x^2}\cos\left(2q\left(x-\dfrac{\alpha}{2}\right)\right)}{2^{2p-1}(p-q)!(p+q)!}dx-\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\sum_{q=0}^p\dfrac{\beta^{2p+1}e^\frac{\alpha^2\beta}{4}\left(x-\dfrac{\alpha}{2}\right)^{2n}e^{-\beta x^2}\cos\left((2q+1)\left(x-\dfrac{\alpha}{2}\right)\right)}{4^p(p-q)!(p+q+1)!}dx$$

$$=\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\dfrac{\beta^{2p}e^\frac{\alpha^2\beta}{4}\left(x-\dfrac{\alpha}{2}\right)^{2n}e^{-\beta x^2}}{4^p(p!)^2}dx+\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\sum_{q=1}^p\dfrac{\beta^{2p}e^\frac{\alpha^2\beta}{4}(\cos\alpha q)\left(x-\dfrac{\alpha}{2}\right)^{2n}e^{-\beta x^2}\cos2qx}{2^{2p-1}(p-q)!(p+q)!}dx+\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\sum_{q=1}^p\dfrac{\beta^{2p}e^\frac{\alpha^2\beta}{4}(\sin\alpha q)\left(x-\dfrac{\alpha}{2}\right)^{2n}e^{-\beta x^2}\sin2qx}{2^{2p-1}(p-q)!(p+q)!}dx-\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\sum_{q=0}^p\dfrac{\beta^{2p+1}e^\frac{\alpha^2\beta}{4}\left(\cos\dfrac{\alpha(2q+1)}{2}\right)\left(x-\dfrac{\alpha}{2}\right)^{2n}e^{-\beta x^2}\cos((2q+1)x)}{4^p(p-q)!(p+q+1)!}dx-\int_{-\infty}^\infty\sum\limits_{p=0}^\infty\sum_{q=0}^p\dfrac{\beta^{2p+1}e^\frac{\alpha^2\beta}{4}\left(\sin\dfrac{\alpha(2q+1)}{2}\right)\left(x-\dfrac{\alpha}{2}\right)^{2n}e^{-\beta x^2}\sin((2q+1)x)}{4^p(p-q)!(p+q+1)!}dx$$

• What a pretty answer:) Thanks, perhaps maybe there is light at the end of this tunnel. +1 – Jeff Faraci Apr 29 '14 at 20:39