# Integral from Cambridge STEP Paper.

• May 26th 2008, 06:41 PM
AAKhan07
Integral from Cambridge STEP Paper.
Show, by means of a suitable change of variable, or otherwise, that

$
\int_0^\infty f((x^2+1)^{1/2} +x) dx = \frac{1}{2}\int_1^\infty (1+t^{-2})f(t) dt
$

Hence, or otherwise, show that

$
\int_0^\infty ((x^2+1)^{1/2} +x)^{-3} dx = \frac{3}{8}
$

This is taken directly from STEP 1, 1998. Now the challenge isn't finding a solution, that's cake for many of you, the challenge is explaining it to me. A maths student in the first year of sixth-form (I think you call it high-school in America.) I've never seen integrals like these before, the most advanced form of integration I've used is integrations by parts.

• May 26th 2008, 07:25 PM
bobak
Quote:

Originally Posted by AAKhan07
Show, by means of a suitable change of variable, or otherwise, that

$
\int_0^\infty f((x^2+1)^{1/2} +x) dx = \frac{1}{2}\int_1^\infty (1+t^{-2})f(t) dt
$

Hence, or otherwise, show that

$
\int_0^\infty ((x^2+1)^{1/2} +x)^{-1/3} dx = \frac{3}{8}
$

This is taken directly from STEP 1, 1998. Now the challenge isn't finding a solution, that's cake for many of you, the challenge is explaining it to me. A maths student in the first year of sixth-form (I think you call it high-school in America.) I've never seen integrals like these before, the most advanced form of integration I've used is integrations by parts.

Pleas explain EVERYTHING!

The questions requires a simple change of variable (which you do learn in high school). you start by letting $t = (x^2+1)^{1/2} +x$ as x becomes large so does t. but for x = 0 t = 1 hence the limits change.

$dt/dx = x(x^2+1)^{-1/2} + 1$ do $dx = \frac{dt}{x(x^2+1)^{-1/2} + 1}$ now you need to show that $\frac{dt}{x(x^2+1)^{-1/2} + 1}$ is the same as $1/2(1+t^{-2})$ and the algebra involves in this is quiet fiddly.

firstly $\frac{1}{x(x^2+1)^{-1/2} + 1} = \frac{(x^2+1)^{1/2}}{x + (x^2+1)^{1/2}} = \frac{(x^2+1)^{1/2}}{t}$ now multiply top and bottom by t $\frac{(x^2+1)^{1/2}( (x^2+1)^{1/2} +x)}{t^2}$ I'm going to ommit a bit of algebra here but you should be able to work it out as $1/2 (\frac{1 + t^2}{t^2}) = 1/2 (1+t^{-2})$.

For the next part you use the pervious result.

let $f = x^{-3}$
you integral is transformed to $1/2 \int_1^\infty (1+t^{-2})(t)^{-3} dt \ \ \Rightarrow 1/2 \int_1^\infty (t^{-3}+t^{-5}) dt$

$= 1/2 \left| -(1/2) t^{-2} - (1/4) t^{-4} \right| _1^\infty = (1/2)( 0 + 1/2 + 1/4 ) = 3/8$

note you don't need any advance knowledge of convergence / divergence to state that $\frac{1}{ \infty} = 0$

Bobak
• May 27th 2008, 06:44 AM
AAKhan07
OK, understood.
It all looks pretty clear, it's going to take a bit of brain training before I can do that kind of work independently, but it's nothing too hard. Thanks Bobak.