## Example 8.3.12 (b): Differentiating and Integrating Power Series

Integrate the series

*f(x) =*twice and simplify your answer. Can you use your result to figure out what function this power series represents?The series converges for all *x* (confirm!) and can therefore
be integrated (and differentiated) term-by-term. We have:

and integrating again:

Now, the above integration is technically correct, but with all the
summation signs, integrals, and indeces it is difficult to see what is
*really* going on. Things get a little clearer if we write out the
series and integrate 'manually':

f(x) = =

= 1 - 1/2! x^{2}+ 1/4! x^{4}- 1/6! x^{6}+ ...

Integrating gives us a new function *g(x)*:

g(x) = f(x) dx = x - 1/3! x^{3}+ 1/5! x^{5}- 1/7! x^{7}+ ... + C

Integrating again gives us:

g(x) dx = 1/2 x^{2}- 1/4! x^{4}+ 1/6! x^{6}- 1/8! x^{8}+ ... + Cx + D =

= -f(x) + 1 + Cx + D' =

= -f(x) + Cx + D''

where we subsume the *+1* into the new constant *D'*.

In other words, if we ignore the constants we have two functions *f*
and *g* such that

f(x) dx = g(x) and g(x) dx = -f(x)

Hmmm ... what well-known functions could *f* and *g* be?
To help your memory, here is the graph of these functions:

f(x) =g(x) =