## 8.3. Series and Power Series

### 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! x2 + 1/4! x4 – 1/6! x6 + …

Integrating gives us a new function g(x):

g(x) =f(x) dx = x – 1/3! x3 + 1/5! x5 – 1/7! x7 + … + C

Integrating again gives us:

g(x) dx = 1/2 x2 – 1/4! x4 + 1/6! x6 – 1/8! x8 + … + 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) =