This integral has one point that makes it improper for any `p , ` namely `+ oo ,` and can have one more such point, namely `x = 0 , ` if `1 / x^p ` is unbounded near `0 .`

There are no more such points because for any `p ` and `x gt 0 ` the function is continuous.

Further, near `x = 0 ` the only part that affects convergence is `1 / x^p ` because `1 / ( 1 + x^3 ) ` is continuous and bounded. The integral `int_0^A 1 / x^p dx ` converges if and only if `p lt 1 .`

Near `x + oo ` the exponent `p ` must at least be positive because for non-positive `p ` the function `gt= 1 ` for `x gt= 1 . ` For a positive `p ` the function tends to zero with the speed equivalent to `1 / x^( 4p ) ` and is positive, so it must be `4p gt 1 ,` i.e. `p gt 1 / 4 .`

In total, for the entire integral to converge, `p ` must be in `( 1 / 4 , 1 ) .`