I've come across this funny problem while messing around with integration by parts. Probably made a mistake somewhere. In the integration of parts expression, it's possible to expand it further. Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image! Plugging the second expression into the first, we get Please Register or Log in to view the hidden image! I don't think this is standard notation, but it's better than writing out all the "violin holes". Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image! Expanding ad infinitum we get Please Register or Log in to view the hidden image! The problem is that it seems like we will get an infinite string of polynomials on the right hand side, but not on the left.
It is quite interesting to experiment with all manner of recursive objects. The problem is to find out why there isn't a problem with the maths.
Integration by parts is not an answer to integration -- it's a strategy that sometime helps find a solution, analogous to walking a tree to find a solution to a problem. The problem is that \(\int w_{\tiny 0} dx\) doesn't automatically tell you how to factor it into \( \int u v' dx = uv - \int v u' dx\) such that \(\int v u' dx = \int w_{\tiny 1} dx \) is closer to an answer than \(\int w_{\tiny 0} dx\) was. Specifically nothing prevents the solution to \(\int w_{\tiny 1} dx\) as also requiring integration by parts. Example: \( \int x^3 \, \sin x \, dx \; = \; - x^3 \, \cos x \; - \; \int -3x^2 \, \cos x \, dx \\ \quad \quad \quad = \; - x^3 \, \cos x \; - \; \left( -3 x^2 \, \sin x \; - \; \int -6x \, \sin x \, dx \right) \\ \quad \quad \quad = \; - x^3 \, \cos x \; - \; \left( -3 x^2 \, \sin x \; - \; \left( 6x \, \cos x \; - \; \int -6 \, \cos x \, dx \right) \right) \\ \quad \quad \quad = - x^3 \, \cos x \; + \; 3 x^2 \, \sin x \; + \; 6x \, \cos x \; - \; 6 \sin x \; + \; C \\ \int w_{\tiny 0} dx = u_{\tiny 0} v_{\tiny 0} - \int w_{\tiny 1} dx = u_{\tiny 0} v_{\tiny 0} - u_{\tiny 1} v_{\tiny 1} + \int w_{\tiny 2} dx = u_{\tiny 0} v_{\tiny 0} - u_{\tiny 1} v_{\tiny 1} + u_{\tiny 2} v_{\tiny 2} - \int u'_{\tiny 2} v_{\tiny 2} dx \\ \begin{eqnarray} w_{\tiny 0} & = & x^3 \, \sin x \quad \quad \quad & u_{\tiny 0} & = & x^3 \quad \quad \quad & v'_{\tiny 0} & = & \sin x \quad \quad \quad & u'_{\tiny 0} & = & 3x^2 \quad \quad \quad & v_{\tiny 0} & = & - \cos x \\ w_{\tiny 1} & = & -3x^2 \, \cos x \quad \quad \quad & u_{\tiny 1} & = & -3x^2 \quad \quad \quad & v'_{\tiny 1} & = & \cos x \quad \quad \quad & u'_{\tiny 1} & = & -6x \quad \quad \quad & v_{\tiny 1} & = & \sin x \\ w_{\tiny 2} & = & -6x \, \sin x \quad \quad \quad & u_{\tiny 2} & = & -6x \quad \quad \quad & v'_{\tiny 2} & = & \sin x \quad \quad \quad & u'_{\tiny 2} & = & -6 \quad \quad \quad & v_{\tiny 2} & = & - \cos x \end{eqnarray}\)
Yeah, so we always hope that \(\int w_{\tiny 1} dx\) will be in a form we know how to integrate. As for the polynomial string, it is quite complicated. They are multiplied by inverse factorial coefficients and the higher order derivatives of u.
