# Changing integration limits from [0,a] to [0, inf)

1. Oct 12, 2012

### PerUlven

The problem statement, all variables and given/known data
This isn't really homework, but a question I came upon when doing my homework.

How can I go from an integral with limits 0 and a:

$\int_0^a f(x) dx$

to something with limits 0 and $\infty$ (still giving the same answer)

$c\int_0^\infty f(u) du$

, where $c$ is some kind of constant.

The attempt at a solution
I've tried substituting $x$ with $tan(\frac{\pi x}{2a})$, since this seems to give the correct limits, but I don't know if it makes any sense, and what to do next (how to find c for example).

2. Oct 12, 2012

### Ray Vickson

You can't, in general. The integral of f(x) for 0 ≤ x ≤ a depends only on the value of f(x) for x between 0 and a, while the integral of f(u) for 0 ≤ u < ∞ depends on the values of f(u) for all positive values of u, including values u > a. The behaviour of f on [0,a] may be essentially unrealated to the behaviour on [a,∞).

Now, of course, if both integrals are finite, you can get c by dividing one by the other, but that is trivial and probably not what you had in mind. You cannot do that at all if the integral over [0,a] is finite while that over [0,∞) is infinite.

RGV

3. Oct 12, 2012

### deep838

I think it is not possible to change limits from some value to infinity for an unkown function.... I can be wrong but I never saw any such thing before

4. Oct 12, 2012

### SammyS

Staff Emeritus
If you make the substitution, $\displaystyle u=\tan\left(\frac{\pi x}{2a}\right)\,,$ then $\displaystyle x=\frac{2a}{\pi }\tan^{-1}(u)\,,$ so that $\displaystyle dx=\frac{2a}{\pi }\frac{1}{1+u^2}du\ .$

This gives: $\displaystyle \int_0^a f(x)\,dx =\frac{2a}{\pi }\int_0^{\infty} f\left(\frac{2a}{\pi }\tan^{-1}(u)\right)\frac{1}{1+u^2}du\,,$ which is probably not what you wanted.

5. Oct 12, 2012

### PerUlven

Thanks a lot guys. This is what happens when I sit and wonder about mathematics and physics into the night!