# Recurrence relation for an integral equation

1. Aug 1, 2011

### aperception

Hi guys, would appreciate any help with this problem. It's from a graduate school entrance examination which I am practicing.

problem statement

When n is a natural number, the indefinite integral $I_n$ is defined as

$I_n=\int\frac{1}{(x^2+a^2)^n}dx$

Here, a is a constant real number, and not equal to zero. Answer the following questions.

1. Express $I_{n+1}$ as a recurrence equation with $I_n$.
2. Derive $I_1$ and $I_2$

where I'm at
I can solve part 2 relatively easily by using the trigonometric substitution $x=a \tan\theta$ which gives the result (feel free to ask for working if you don't believe me):

$I_n=\frac{1}{a^{2n-1}}\int(\cos\theta)^{2n-2}d\theta$

or:

$I_{n+1}=\frac{1}{a^{2n+1}}\int(\cos\theta)^{2n} d\theta$

these are fairly easy to solve for fixed values of n using basic integration and backwards substitutions such as $\tan\theta=(x/a)$ etc, giving:

$I_1=\arctan(x/a)/a$

$I_2=\arctan(x/a)/(2a^3)+\frac{x}{2a^2(x^2+a^2)}$

but I don't know how to generalize to the final recurrence equation.

attempted solution
I tried to solve the original equation using integration by parts by writing:

$I_{n+1}=\int\frac{1}{(x^2+a^2)^n}\frac{1}{(x^2+a^2)}dx$

then setting:
$u=\frac{1}{(x^2+a^2)}$, $v'=\frac{1}{(x^2+a^2)^n}$

$u'=\frac{-2x}{(x^2+a^2)^2}$, $v=I_n$

this gives:
$I_{n+1}=\frac{I_n}{(x^2+a^2)}+\int\frac{I_n 2x}{(x^2+a^2)^2}dx$

but this doesn't agree with the calculations from above with fixed n, so it must be wrong. (I presume $v=I_n$ is invalid for some reason?).

Unfortunately I don't really have any ideas on where to go from here...?

2. Aug 1, 2011

### SammyS

Staff Emeritus
Try the following: (I haven't checked it out myself, so it may be bogus.)

$\displaystyle I_n=\int\frac{1}{(x^2+a^2)^n}dx$
$\displaystyle =\int\frac{x^2+a^2}{(x^2+a^2)^{(n+1)}}dx$

$\displaystyle =a^2I_{n+1}+\int (x)\,\frac{x}{(x^2+a^2)^{(n+1)}}dx$

Now try integration by parts: u = x, $\displaystyle dv=\frac{x}{(x^2+a^2)^{(n+1)}}dx$​

Last edited: Aug 1, 2011
3. Aug 1, 2011

### aperception

That is very clever, it worked great thank you!