<prompt> Kwa kuzingatia ukweli kwamba int_-inftyinfty e-x2 d xsqrtpi, tathmini muunganisho.Iint_-inftyinfty int_-inftyinfty e-leftx2y-x2y 2kulia d x d y .  <response>: TunaandikaanzailiyopangwaMimi int_-inftyinfty e-3 y2 2leftint_-inftyinfty e-2 x 22 x y-y2 2 d xkulia d y nint_-inftyinfty e-3 y2 2kushoto int_-inftyinfty e-2leftx-fracy2kulia2 d xkulia d ymwishoiliyopangwakufanya ubadilishanaji tsqrt2leftx-fracy2kulia, d tsqrt2 d x kwenye kiungo cha ndani, tunapataanzailiyopangwaMimi frac1sqrt2 int_-inftyinfty e-3 y2 2leftint_-infty infty e-t2 d tkulia d y nsqrtfracpi2 int_-inftyinfty e -3 y2 2 sikumwishoiliyopangwasasa tunabadilisha: ssqrtfrac32 y, d ssqrtfrac32 d y tunapataanzailiyopangwaMimi sqrtfracpi3 int_-inftyinfty e-s2 d s nfracpi sqrt3 .mwishoiliyopangwa 