The following example is very artificial and therefore not quite satisfying, but it answers you question anyway.

Consider . You can check that , so that you may compute using 3 integrations by parts, which will end up with the initial integral.

In fact, if you know complex numbers, is the real part of where is a cubic root of the unity: , so what I claimed about is a simple consequence of this property.

Of course however, since , two integrations by parts would suffice if you split into and . I don't know of a genuine example where after 3 integrations by parts, and not possibly less, we recover the initial integral, in a way enabling to find its value.