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Gaussian Integral Substitution

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pcvrx560
#1
Feb17-13, 03:07 AM
P: 8
If I had an integral

[tex] \int_{-1}^{1}e^{x}dx [/tex]

Then performing the substitution [itex] x=\frac{1}{t} [/itex] would give me

[tex] \int_{-1}^{1}-e^\frac{1}{t}t^{-2}dt [/tex]

Which can't be right because the number in the integral is always negative. Is this substitution not correct?

Sorry if I am being very thick but I can't figure out why I can't evaluate this simple integral with this change of variables.
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tiny-tim
#2
Feb17-13, 05:11 AM
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hi pcvrx560!
Quote Quote by pcvrx560 View Post
Is this substitution not correct?
the substitution is fine, but the limits are wrong

as x goes up from -1 to 1,

t (= 1/x) goes down from -1 to -∞, and then from +∞ down to 1

you'ld need to write ##\int_{-1}^{-\infty} + \int_{\infty}^{1}##
(or ##-\int^{-1}_{-\infty} - \int^{\infty}_{1}##)
SteamKing
#3
Feb17-13, 07:15 AM
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PF Gold
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Isn't the integral of e^x w.r.t. x simply e^x + C?

pcvrx560
#4
Feb17-13, 10:54 PM
P: 8
Gaussian Integral Substitution

Thanks, tiny-tim! That cleared it up for me.

I didn't think about the range I was integrating over, I was just mindlessly plugging numbers into 1/t.


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