# Thread: Integral from Cambridge STEP Paper.

1. ## 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.

2. 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

3. ## 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.