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.
Best Answer
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:
$$ \varphi(x) = c + 1 - \frac{1}{x} $$ $$ \int_{-\infty}^{c} f(x) dx = \int_0^1 f\left(c + 1 - \frac{1}{x} \right) \frac{1}{x^2} dx$$ $$ \varphi(x) = c + \log(x) $$ $$ \int_{-\infty}^{c} f(x) dx = \int_0^1 f(c + \log(x)) \frac{1}{x} dx$$