- #1
geoduck
- 258
- 2
When I type in this integral into Mathematica:
$$\int_1^\infty \int_1^\infty \frac{dxdy}{(x+y^2)^2}$$
using a large number like 1020 instead of ∞, Mathematica gives me 0.785398. No matter what large number I use, Mathematica always gives me around that value.
However, doing the integral by hand, I get it diverges. I make the substitution u=x+y^2 and v=y. The Jacobian of the transformation is 1, so I get:
$$\int_1^\infty \int_1^\infty \frac{dxdy}{(x+y^2)^2}=
\int_1^\infty dv \int_2^\infty \frac{du}{u^2}=(\infty-1)(\frac{1}{2})
$$
which is divergent.
Also, I noticed weird behavior by Mathematica too. If I change the exponent ##(x+y^2)^2## to a non-integer like ##(x+y^2)^{2.1}##, then the integral jumps from being zero or very large.
Does anyone know what's going on here?
$$\int_1^\infty \int_1^\infty \frac{dxdy}{(x+y^2)^2}$$
using a large number like 1020 instead of ∞, Mathematica gives me 0.785398. No matter what large number I use, Mathematica always gives me around that value.
However, doing the integral by hand, I get it diverges. I make the substitution u=x+y^2 and v=y. The Jacobian of the transformation is 1, so I get:
$$\int_1^\infty \int_1^\infty \frac{dxdy}{(x+y^2)^2}=
\int_1^\infty dv \int_2^\infty \frac{du}{u^2}=(\infty-1)(\frac{1}{2})
$$
which is divergent.
Also, I noticed weird behavior by Mathematica too. If I change the exponent ##(x+y^2)^2## to a non-integer like ##(x+y^2)^{2.1}##, then the integral jumps from being zero or very large.
Does anyone know what's going on here?