Csdrhrt

I found a fantastic and generalizable substitution technique for computing definite integrals that go to infinity from either negative infinity or a constant, regardless of the function (sorry for the external link):

http://ab-initio.mit.edu/wiki/index.php/Cubature#Infinite_intervals

But what's killing me is that I need the same sort of thing for a function going from negative infinity to a constant, and I can't convince myself that some obvious transformation of either of these two is correct. Doing u-substitution on any given integrand to take care of a (negative) infinite bound is simple enough, but a general transformation rule like one of the ones in the link - totally independent of the function itself - would be invaluable.

Edit 1 (Might be a Wild Goose Chase)

So the second formula up in that link, is:

$int_{-infty}^{infty}f(x)dx=int_{-1}^1 f(frac{t}{1-t^2})frac{1+t^2}{(1-t^2)^2}dt$

If $a$ represents our generic real-number constant, the above could certainly be broken up as:

$int_{-infty}^{infty}f(x)dx=int_{-1}^a f(frac{t}{1-t^2})frac{1+t^2}{(1-t^2)^2}dt+int_{a}^1 f(frac{t}{1-t^2})frac{1+t^2}{(1-t^2)^2}dt$

Which tempts me to jump to the conclusion that the first term on the RHS can be used as our general negative-infinity-to-constant formula:

$int_{-infty}^{a}f(x)dx=int_{-1}^a f(frac{t}{1-t^2})frac{1+t^2}{(1-t^2)^2}dt$

In order for this to be true, however, the second RHS term would have to complement it by being a valid formula to integrate from any constant up to infinity, IE:

$int_{a}^{infty}f(x)dx=int_{a}^1 f(frac{t}{1-t^2})frac{1+t^2}{(1-t^2)^2}dt$

...but the first formula given at the above link is also supposed to be such a formula, solving for that exact LHS term! So the two would have to be equivalent - IE:

$int_{a}^{infty}f(x)dx=int_{a}^1 f(frac{t}{1-t^2})frac{1+t^2}{(1-t^2)^2}dt=int_{0}^1 f(a+frac{t}{1-t})frac{1}{(1-t)^2}dt$

And here I'm stuck, because I can't figure out whether or not that's plausible for all convergent $f(x)$. I have a bad feeling about that whole approach because it relies on the assumption that dividing up a finite space at a given point is the same as dividing up an equivalent infinite space at the exact same point, proportionally.

From the wikipedia page for integration by substitution, you need a differentiable function $varphi$ with integrable derivative. If $f$ is continuous, then

$$ int_{varphi(a)}^{varphi(b)} f(x) dx = int_a^b f(varphi(x)) varphi'(x) dx$$

As long as $varphi(a) = -infty$ and $varphi(b) = c$, it should work.
Some functions you can use are:

$$ varphi(x) = c + 1 - frac{1}{x} $$
$$ int_{-infty}^{c} f(x) dx = int_0^1 fleft(c + 1 - frac{1}{x} right) frac{1}{x^2} dx$$
$$ varphi(x) = c + log(x) $$
$$ int_{-infty}^{c} f(x) dx = int_0^1 f(c + log(x)) frac{1}{x} dx$$