# Symbolic Integration

Properties of Symbolic Integration mirror those of Symbolic Differentiation.

#### Linearity

Basic linearity properties of the integral $F(t)=\int_0^tf(s)\, ds$ follow directly from linearity of the underlying initial value problem DF = f with D = d/dt:

• $\int_0^t(f(s)+g(s))\, ds=\int_0^tf(s)\, ds+\int_0^tg(s)\, ds$,
• $\int_0^t\alpha f(s)\, ds=\alpha\int_0^tf(s)\, ds$,

where $\alpha\in R$ is a constant. Further, for a < b < c,

• $\int_a^bf(s)\, ds+\int_b^cf(s)\, ds=\int_a^cf(s)\, ds$,

which is extended to arbitrary limits a, b and c and by defining for b < a.

• $\int_a^bf(s)\, ds=-\int_b^af(s)\, ds$.

Alternatively, these rules can be derived directly from the Riemann-sum representation of the integral.

#### Integration by Parts 1d

We have using the product rule for differentiation with D = d/ds

• $F(t)G(t)-F(0)G(0)=\int_0^tD(F*G)ds =\int_0^t(DF*G+F*DG)ds$

which can be written

• $\int_0^tDF*G\, ds = [F*G]_0^t - \int_0^tF*DG, ds$,

with $[F*G]_0^t = F(t)*G(t)-F(0)*G(0)$. We see that we can “move” the  derivative D from F to G, if we change sign and take into account the difference of end-point values of F*G.

#### Change of Integration Variable

Let H(s) = F(g(s)) be a composite function and recall the Chain Rule $\frac{dH}{ds} = \frac{dF}{dg}*\frac{dg}{ds}$, to get

• $\int_0^t\frac{dH}{ds}\,ds = \int_0^t\frac{dF}{dg}*\frac{dg}{ds}\,ds = \int_{g(0)}^{g(t)}\frac{dF}{dg}*dg$,

with $dg = \frac{dg}{ds}*ds$. We can view this as a “change of integration variable” from s to g with change of time step from ds to dg connected by $dg = \frac{dg}{ds}*ds$ and corresponding change of limits of integration.

#### Integration by Parts 2d and FEM

Recall the Divergence Theorem:

• $\int_D \nabla\cdot q\,dxdy =\int_B q\cdot n\, ds$.

where D domain with boundary B with outward normal $n = (nx,ny), q = (qx,qy), q\cdot n = qx*nx + qy*ny, \nabla\cdot q = \frac{\partial qx}{\partial x}+ \frac{\partial qy}{\partial y}$.

Apply to $q =v\nabla u$ with $\nabla u = (\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y})$ to get

• $\int_D(\nabla v\cdot\nabla u + v\Delta u)\, dxdy =\int_Bv\frac{\partial u}{\partial n}\, ds$

where

• $\nabla v\cdot\nabla u = \frac{\partial v}{\partial x}\frac{\partial u}{\partial x} +\frac{\partial v}{\partial y}\frac{\partial u}{\partial y}$
• $\frac{\partial u}{\partial n}=\frac{\partial u}{\partial x}nx+\frac{\partial u}{\partial y}ny$

In particular, if v=0 on B:

• $\int_D\nabla v\cdot\nabla u\, dxdy = -\int_D v\Delta u\, dxdy.$

This is used  to reformulate the Poisson problem -Δ u = f in D with u=0 on B according to the finite element method FEM into finding $u_h\in V_h$ such that for all $v\in V_h$

• $\int_D\nabla v\cdot\nabla u\, dxdy =\int_D v*f\, dxdy$,

where V_h is a space of continuous piecewise linear functions defined on a triangulation of D which vanish on B. On June 30 2007 The Swedish Parliament dismantled Integrationsverket, the Ministry for Integration, and replaced it by the Ministry for Time-Stepping.