计算无界区域的二重积分∫∫e^[-(x^2+y^2)][sin(x^2+y^2)]^ndxdy其中积分区域D是整个平面,n=1,2,3...注：sin(x^2+y^2)是和e相乘的,不是e的幂数希望给出地推公式的推导后，有进一步的过程，我只计算到n

Question

计算无界区域的二重积分
∫∫e^[-(x^2+y^2)][sin(x^2+y^2)]^n dxdy
其中积分区域D是整个平面,n=1,2,3...
注：sin(x^2+y^2)是和e相乘的,不是e的幂数
希望给出地推公式的推导后，有进一步的过程，我只计算到n和n-2地推公式，然后就不知道改怎么办了。。。。
一楼的回答，没怎么看懂。。。。。

Answer 1

∫∫e^[-(x^2+y^2)][sin(x^2+y^2)]^n dxdy
其中积分区域D是整个平面,n=1,2,3...
x = rcosa,y = rsina,dxdy = rdrda,
x,y:-无穷大 -> +无穷大,
r:0 -> +无穷大,
a:0 -> 2PI.
f(n) = ∫∫_{D}e^[-(x^2+y^2)][sin(x^2+y^2)]^n dxdy
= ∫_{0}^{2PI}da∫_{0}^{+无穷大}e^(-r^2)[sin(r^2)]^nrdr
= PI∫_{0}^{+无穷大}e^(-r)[sin(r)]^ndr
f(1) = PI∫_{0}^{+无穷大}e^(-r)[sin(r)]dr
= PI{-e^(-r)(sinr)}|_{0}^{+无穷大} + PI∫_{0}^{+无穷大}e^(-r)cosrdr
= PI∫_{0}^{+无穷大}e^(-r)cosrdr
= PI{-e^(-r)(cosr)}|_{0}^{+无穷大} - PI∫_{0}^{+无穷大}e^(-r)sinrdr
= PI - f(1)
f(1) = PI/2,
f(2) = PI∫_{0}^{+无穷大}e^(-r)[sin(r)]^2dr
= PI{-e^(-r)(sinr)^2}|_{0}^{+无穷大} + 2PI∫_{0}^{+无穷大}e^(-r)(sinr)cosrdr
= 2PI∫_{0}^{+无穷大}e^(-r)(sinr)cosrdr
= PI{-e^(-r)sin(2r)}|_{0}^{+无穷大} - PI∫_{0}^{+无穷大}e^(-r)[2cos(2r)]dr
= -2PI∫_{0}^{+无穷大}e^(-r)[cos(2r)]dr
= -2PI∫_{0}^{+无穷大}e^(-r)[1 - 2(sinr)^2]dr
= -2PI∫_{0}^{+无穷大}e^(-r)dr + 4PI∫_{0}^{+无穷大}e^(-r)[(sinr)^2]dr
= 2PI{e^(-r)}|_{0}^{+无穷大} + 4f(2)
= -2PI + 4f(2),
f(2) = 2PI/3
n>2时,
f(n) = PI∫_{0}^{+无穷大}e^(-r)[sin(r)]^ndr
= PI{-e^(-r)(sinr)^n}|_{0}^{+无穷大} + nPI∫_{0}^{+无穷大}e^(-r)(sinr)^(n-1)cosrdr
= nPI∫_{0}^{+无穷大}e^(-r)(sinr)^(n-1)cosrdr
= nPI{-e^(-r)(sinr)^(n-1)cosr}|_{0}^{+无穷大} + nPI∫_{0}^{+无穷大}e^(-r)[(n-1)(sinr)^(n-2)(cosr)^2 - (sinr)^n]dr
= nPI∫_{0}^{+无穷大}e^(-r)[(n-1)(sinr)^(n-2)(cosr)^2 - (sinr)^n]dr
= nPI∫_{0}^{+无穷大}e^(-r)[(n-1)(sinr)^(n-2) - n(sinr)^n]dr
= n(n-1)f(n-2) - n^2f(n),
f(n) = n(n-1)f(n-2)/[1+n^2].
因此,
f(1) = PI/2,
f(2) = 2PI/3,
f(n) = n(n-1)f(n-2)/[1+n^2],n = 3,4,...
比如,
f(3) = 3(3-1)f(1)/[1+3^2] = 3PI/10
f(4) = 4(4-1)f(2)/[1+4^2] = 8PI/17,
...