178 3 Real Analysis
∫b
a
∑∞
n= 0
f (n, x)=
∑∞
n= 0
∫b
a
f (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=
∫∞
0
1
√
x
e−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=
∫∞
0
1
√
x
e−xdx=
1
√
π
∫∞
0
∫∞
−∞
e−xt
2
e−xdtdx=
1
√
π
∫∞
−∞
∫∞
0
e−(t
(^2) + 1 )x
dxdt
=
1
√
π
∫∞
−∞
1
t^2 + 1
dt=
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,
∫∞
0
e−sxx−^1 sinxdx=arctan(s−^1 ).