152 3 Real Analysis
Proof.To prove the formula we start by computing recursively the integral
In=∫ π 40tan^2 nxdx, n≥ 1.We have
In=∫ π 40tan^2 nxdx=∫ π 40tan^2 n−^2 xtan^2 xdx=
∫ π 40tan^2 n−^2 x( 1 +tan^2 x)dx−∫ π 40tan^2 n−^2 xdx=
∫ π 40tan^2 n−^2 xsec^2 xdx+In− 1.The remaining integral can be computed using the substitution tanx=t. In the end, we
obtain the recurrence
In=1
2 n− 1−In− 1 ,n≥ 1.So forn≥1,
In=1
2 n− 1−
1
2 n− 3+···+
(− 1 )n−^2
3
+(− 1 )n−^1 I 1 ,with
I 1 =
∫ π 40tan^2 xdx=∫ π 40sec^2 xdx−∫ π 401 dx=tanx∣
∣∣π 4
0 −π
4= 1 −
π
4.
We find that
In=1
2 n− 1−
1
2 n− 3+···+
(− 1 )n−^2
3+(− 1 )n−^1 +(− 1 )n
π
4.
Because tan^2 nx→0asn→∞uniformly on any interval of the form[ 0 ,a),a<π 4 ,it
follows that limn→∞In=0. The Leibniz formula follows.
Below are more examples of this kind.464.LetP(x)be a polynomial with real coefficients. Prove that
∫∞
0e−xP(x)dx=P( 0 )+P′( 0 )+P′′( 0 )+···.