Z Question IN (1+ n I
Question Find a reduction formula for IN =
0
(1
dx + x2)n
.
Work out
the
value of I1.
Hence evaluate this integral for arbitrary positive integer n,
Answer In =
dx 0 (x2 + 1)n
(Assume it converges for n positive and > 1)
Integrate by parts with
u
=
1 (1 + x2)n
du dx
=
-n ? 2x (1 + x2)n+1
dv = 1 dx v=x
Therefore
In
= =
x (1 + x2)n
-
0
0
dxx
?
(-2nx) (1 + x2)n+1
0(if n > 1) + 2n
dxx2 0 (1 + x2)n+1
So In = 2n
dxx2 0 (1 + x2)n+1
What now? Use a similar trick to Q3.
x2 = (1 + x2) - 1
So
In
=
2n
0
dx[(1 + x2) - 1] (1 + x2)n+1
=
2n
0
dx(1 + x2) (1 + x2)n+1
-
2n
dx 0 (1 + x2)n+1
=
2n
0
(1
dx + x2)n
-2n
dx
0 (1 + x2)n+1
In
In+1
Therefore In = 2nIn - 2nIn+1
or
In+1
=
(2n - 2n
1) In
This can be rewritten by putting n + 1 n
or
In
=
(2(n - 1) - 2(n - 1)
1) In-1
In =
2n - 3 2n - 2
In-1
()
1
What's I1?
I1 =
0
dx (1 + x2)1
=
0
(1
dx + x2)
=
[arctan x] 0
+
2
Hence applying ( ) recursively:
In =
2n - 3 2n - 2
In-1
In-1 =
2n - 5 2n - 4
In-2
In-2 =
2n - 7 2n - 6
In-3
In-3 = .........etc.
......
I2 =
2?2-3 2?2-2
I1
=
1 22
Recursion. This assumes that n is a positive integer.
Hence
In
=
(2n (2n
- -
3) 2)
?
(2n (2n
- -
5) 4)
?
(2n (2n
- -
7) 6)
?
???
?
1 2
?
2
2
................
................
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