13.1 Iterated integrals 653
Substituting this in the formula for f2(x) then gives
(13.12) .12(x) = fax (fatf(ti) dti) dt.
Replacing ti by t2 and then t by ti and x by t gives
f2 (t) =f t ( at`a f(t2) dtZ) dti
and substituting this into the formula for f3(x) gives
(13.13)
f3(x) = fax l ft (fats f(ts)
dt2) dtl} dt.
The integrals in (13.12) and (13.13) are examples of iterated integrals
and, for each n, we could write a formula for ff(x) which involvesn of
these integrals.
In case f(x) = 1 for each x, we do not need the iterated integralsto
obtain formulas for fi(x), f2(x), ; we can use the formulas (13.11)
one after another to obtain
(13.14) fi(x) = x - a, f2(x) _ (x - a)f3(x) _(x - a)
2! 3!
f4(x) _ (x
41
a)4
We must, however, learn how iterated integrals are manipulated when
they appear in our work and cannot be avoided. The way in which
(13.12) was obtained tells us that to get f2(x) we should integrate first
with respect to ti to evaluate the integral in parentheses to obtain a
function of t which is integrated with respect to t to obtain f2(x). A
tempest in a teapot appears when we, like everyone else, adopt the view
that parentheses are nuisances and write (13.12) in the form
(13.15) f2(x) = f ax ftf(ti) dti dt
and insist that we must find f2(x) by integrating first with respect to ti,
the limits of integration being a and t, and integrating last with respect
to t from a to x. The difficulty lies in the fact that there is always the
possibility of constructing a theory of iterated integrals in such a way
that, for example,
(13.16) f ' f D F(x,y) dx dy
means the result of integrating first with respect to y from C to D (not
from A to B) and then integrating last with respect to x from A to B.
It is equally sensible to insist on one hand that we should "work outward
from the middle," so that fc goes with dx and f4 goes with dy, and toD