how do you show that intergrate from o to infinity of exp (-r^2) over (r^4 + a^4) is less than pi over ( (2)^(0.5) a^3)
given that integrate from o to infinity of 1/ (x^2 +a^2 ) dx is pi over (2a)
Since $\displaystyle e^{-r^2}<1$ whenever r>0, it follows that $\displaystyle \int_0^\infty\!\!\frac{e^{-r^2}}{r^4+a^4}\,dr < \int_0^\infty\!\!\frac1{r^4+a^4}\,dr $. Substitute r=ax into that second integral to see that it is equal to $\displaystyle \frac1{a^3}\int_0^\infty\!\!\frac1{1+x^4}\,dx$. The integral $\displaystyle \int_0^\infty\!\!\frac1{1+x^4}\,dx$ can be evaluated (by contour integration, or by partial fractions), and it is equal to $\displaystyle \frac{\pi}{2\sqrt2}$.
Putting that all together, you get $\displaystyle \int_0^\infty\!\!\frac{e^{-r^2}}{r^4+a^4}\,dr < \frac{\pi}{2a^3\sqrt2}$. That is better than what was asked for, because it has an extra 2 in the denominator. But I don't see any way of getting this result by using the given hint about $\displaystyle \int_0^\infty\!\!\frac1{x^2+a^2}\,dx$.