178 3 Real Analysis
∫ba∑∞
n= 0f (n, x)=∑∞
n= 0∫baf (n, x).Here we are allowed to commute the sum and the integral if eitherfis a positive function,
or if
∫b
a∑∞
n= 0 |f (n, x)|(or equivalently∑∞
n= 0∫b
a|f (n, x)|) is finite. It is now time for
an application.
Example.Compute the integral
I=
∫∞
01
√
xe−xdx.Solution.We will replace√^1 xby a Gaussian integral. Note that forx>0,
∫∞−∞e−xt
2
dt=∫∞
−∞e−(√xt) 2
dt=1
√
x∫∞
−∞e−u
2
du=√
π
x.
Returning to the problem, we are integrating the positive function√^1 xe−x, which is inte-
grable over the positive semiaxis because in a neighborhood of zero it is bounded from
above by√^1 xand in a neighborhood of infinity it is bounded from above bye−x/^2.
Let us consider the two-variable functionf (x, y)=e−xt
2
e−x, which is positive and
integrable overR×( 0 ,∞). Using the above considerations and Tonelli’s theorem, we
can write
I=∫∞
01
√
xe−xdx=1
√
π∫∞
0∫∞
−∞e−xt
2
e−xdtdx=1
√
π∫∞
−∞∫∞
0e−(t(^2) + 1 )x
dxdt
=
1
√
π∫∞
−∞1
t^2 + 1dt=π
√
π=
√
π.Hence the value of the integral in question isI=
√
π. More applications are given below.521.Leta 1 ≤a 2 ≤ ··· ≤an=mbe positive integers. Denote bybkthe number of
thoseaifor whichai≥k. Prove that
a 1 +a 2 +···+an=b 1 +b 2 +···+bm.522.Show that fors>0,
∫∞
0e−sxx−^1 sinxdx=arctan(s−^1 ).