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If I have the integral: $\displaystyle\int_{0}^{\infty}t^{-\frac{1}{2}}e^{-t} dt$

Am I allowed to make the substitution $t=x^2$, because I am then not sure what the limits of integration would be as for $t$ positive $x$ could be negative or positive?

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  • $\begingroup$ so you don't know how to do a change of variable : you need to consider a function such that it is bijective $C^1$ on the interval of integration... of course $x = \sqrt{t}, t = x^2$ is bijective on $[0,\infty[$ $\endgroup$ – reuns May 22 '16 at 22:41
  • $\begingroup$ More than bijective, it is increasing. $0$ maps to $0$ and $\infty$ maps to $\infty$. $\endgroup$ – GEdgar May 22 '16 at 23:11
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Hint. You may rather perform the change of variable $x=\sqrt{t}>0$, giving $dx=\frac{1}{2}t^{-\frac{1}{2}}dt$ to get $$ \int_{0}^{\infty}t^{-\frac{1}{2}}e^{-t} dt=2\int_{0}^{\infty}e^{-x^2} dx $$ then you may use the standard gaussian result.

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  • $\begingroup$ This is the same change of variable as the one the OP is suggesting. $\endgroup$ – GFauxPas May 22 '16 at 21:32
  • $\begingroup$ Not exactely. The change $x=\sqrt{t}$ is a one to one function ( $(0,\infty)$ and $(0,\infty)$) whereas $t=x^2$ (the OP's) is not. $\endgroup$ – Olivier Oloa May 22 '16 at 21:40
  • $\begingroup$ you do not need the change to be injective in the one dimensional case. $\endgroup$ – GFauxPas May 24 '16 at 17:30
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Here are the conditions for making a substitution:

$\int_{y_0}^{y_1}f(y) dy = \int_{x_0}^{x_1}f(h(x))h'(x) dx$

if: 1. $f : [y_0, y_1] → R$ is continuous on $[y_0, y_1]$,

  1. $h : [x_0, x_1] → [y_0, y_1]$ is differentiable on $[x_0, x_1]$, with $h ′$ continuous on $[x_0, x_1]$,
  2. $h(x_0) = y_0$ and $h(x_1) = y_1$

note that this doesn't actually mean that h has to be bijective.

By taking limits you can extend this to improper integrals with infinities in the limits.

Also if you can satisfy the conditions 1 and 2 above but $h(x_0)=y_1$ and $h(x_1)=y_0$ ($y_1 > y_0$) you can fiddle about and show that you still $\int_{y_0}^{y_1}f(y) dy = \int_{x_0}^{x_1}f(h(x))h'(x) dx$ (do this by subtituting with $h_2(x) = h(x_0 + x_1 - x)$ which does satisfy the conditions above).

So you can actually get it to work both ways,

In your case:

$t(x) = x^2$

$t((- \infty ,0]) = [0, \infty)$

$t'(x) = 2x$

$\int_{0}^{\infty}t^{-\frac{1}{2}}e^{-t} dt = \int_{0}^{-\infty}|x^{-1}|e^{-x^2}*2x dx = \int_{0}^{-\infty}-2e^{-x^2} dx$ (x<=0)

$=\int_{-\infty}^{0}2e^{-x^2} dx$

And you can see that this is going to be the same as $\int_{0}^{\infty}2e^{-x^2} dx$ which is what you get if you do it with the other choice of limits.

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  • $\begingroup$ to finish off there's a nice trick involving polar coordinates which you can probably find easily on the internet $\endgroup$ – Luke Naylor May 22 '16 at 22:52

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