# Revision history [back]

Here comes a possible solution using (directly) the numerical_integral. (Without getting errors.)

First, let us introduce the function $$h(x,y) =\frac 1{2\pi\cdot(1+|1+x+y|^3)}\exp-\frac {x^2+y^2}2\ .$$ We will immediately also compute numerically the integral $$\int_{-20}^{20} h(0,y)\; dy\ ,$$ because the (type for the) sage result is possibly unexpected.

sage: h(x,y) = exp( -(x^2+y^2)/2 ) / 2 / pi / ( 1+abs(x+y+1)^3 )
sage: numerical_integral( lambda y : h(0,y), (-20,20) )
(0.2032706496721935, 2.562242049607144e-09)
sage: numerical_integral( lambda y : h(0,y), (-20,20) )[0]
0.2032706496721935

So the expression numerical_integral( lambda y : h(0,y), (-20,20) ) is not a float number, but a tuple containing an approximative value for the integral, and the error made during the computation.

So a possible way to get the integral is...

sage: h(x,y) = exp( -(x^2+y^2)/2 ) / 2 / pi / ( 1+abs(x+y+1)^3 )
sage: numerical_integral( lambda x : numerical_integral( lambda y : h(x,y), (-20,20) )[0], (-20,20) )
(0.45483611545273767, 1.9556993802183342e-09)

This is in concordance with gp/pari, for instance:

? \p 50
realprecision = 57 significant digits (50 digits displayed)
? h(x,y) = exp( -(x^2+y^2)/2 ) / 2 / Pi / ( 1 + abs(x+y+1.0)^3 )
%1 = (x,y)->exp(-(x^2+y^2)/2)/2/Pi/(1+abs(x+y+1.0)^3)
? intnum( x=-20,20, intnum( y=-20,20, h(x,y) ) )
%2 = 0.45483723187162310868693510119720179912113863975808

Note: The given integral is roughly the expected value of the random variable $$\frac 1{1+|1+X+Y|^3}\ ,$$ where $X,Y$ are two independent standard normal random variables. (Distributed following the $N(0,1^2)$ law...)

(We integrate on the full real line instead of $[-20,+20]$, the exponential decay is already strong.)

We can of course change variables using instead $Z,W=\frac 1{\sqrt 2}(X\pm Y)$. (Use the plus for $Z$, the minus for $W$.) The corresponding passage matrix is orthogonal. So the integral should be the same as $${\mathbb E}\left[\ \frac 1{1+|1+X+Y|^3}\ \right]={\mathbb E}\left[\ \frac 1{1+|1+Z\sqrt 2|^3}\ \right]=\int_{-\infty}^{\infty}\frac 1{1+|1+z\sqrt 2|^3}\; \frac 1{\sqrt{2\pi}}\; \exp -\frac {z^2}2\; dz\ .$$ One more numerical check:

sage: numerical_integral( 1/(1+abs(1+z*sqrt(2))^3) * exp(-z^2/2), [-20,20] )[0] / sqrt(2*pi).n()
0.454836900304870
sage: numerical_integral( 1/(1+abs(1+z*sqrt(2))^3) * exp(-z^2/2) / sqrt(2*pi), [-20,20] )
(0.4548369003048696, 5.4312008620095026e-08)

Here comes a possible solution using (directly) the numerical_integral. (Without getting errors.)

First, let us introduce the function $$h(x,y) =\frac 1{2\pi\cdot(1+|1+x+y|^3)}\exp-\frac {x^2+y^2}2\ .$$ We will immediately also compute numerically the integral $$\int_{-20}^{20} h(0,y)\; dy\ ,$$ because the (type for the) sage result is possibly unexpected.

sage: h(x,y) = exp( -(x^2+y^2)/2 ) / 2 / pi / ( 1+abs(x+y+1)^3 )
sage: numerical_integral( lambda y : h(0,y), (-20,20) )
(0.2032706496721935, 2.562242049607144e-09)
sage: numerical_integral( lambda y : h(0,y), (-20,20) )[0]
0.2032706496721935

So the expression numerical_integral( lambda y : h(0,y), (-20,20) ) is not a float number, but a tuple containing an approximative value for the integral, and the error made during the computation.

So a possible way to get the integral is...

sage: h(x,y) = exp( -(x^2+y^2)/2 ) / 2 / pi / ( 1+abs(x+y+1)^3 )
sage: numerical_integral( lambda x : numerical_integral( lambda y : h(x,y), (-20,20) )[0], (-20,20) )
(0.45483611545273767, 1.9556993802183342e-09)

This is in concordance with gp/pari, for instance:

? \p 50
realprecision = 57 significant digits (50 digits displayed)
? h(x,y) = exp( -(x^2+y^2)/2 ) / 2 / Pi / ( 1 + abs(x+y+1.0)^3 )
%1 = (x,y)->exp(-(x^2+y^2)/2)/2/Pi/(1+abs(x+y+1.0)^3)
? intnum( x=-20,20, intnum( y=-20,20, h(x,y) ) )
%2 = 0.45483723187162310868693510119720179912113863975808

Note: The given integral is roughly the expected value of the random variable $$\frac 1{1+|1+X+Y|^3}\ ,$$ where $X,Y$ are two independent standard normal random variables. (Distributed following the $N(0,1^2)$ law...)

(We integrate on the full real line instead of $[-20,+20]$, the exponential decay is already strong.)

We can of course change variables using instead $Z,W=\frac 1{\sqrt 2}(X\pm Y)$. (Use the plus for $Z$, the minus for $W$.) The corresponding passage matrix is orthogonal. So the integral should be the same as $${\mathbb E}\left[\ \frac 1{1+|1+X+Y|^3}\ \right]={\mathbb E}\left[\ \frac E}\left[\frac 1{1+|1+X+Y|^3}\right]={\mathbb E}\left[\frac 1{1+|1+Z\sqrt 2|^3}\ \right]=\int_{-\infty}^{\infty}\frac 2|^3}\right]=\int_{\mathbb R}\frac 1{1+|1+z\sqrt 2|^3}\; \frac 1{\sqrt{2\pi}}\; \exp -\frac {z^2}2\; dz\ .$$ One more numerical check:

sage: numerical_integral( 1/(1+abs(1+z*sqrt(2))^3) * exp(-z^2/2), [-20,20] )[0] / sqrt(2*pi).n()
0.454836900304870
sage: numerical_integral( 1/(1+abs(1+z*sqrt(2))^3) * exp(-z^2/2) / sqrt(2*pi), [-20,20] )
(0.4548369003048696, 5.4312008620095026e-08)