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Gaussian Integral Substitution |
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| Feb17-13, 03:07 AM | #1 |
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Gaussian Integral Substitution
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. |
| Feb17-13, 05:11 AM | #2 |
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hi pcvrx560!
![]() 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}##) |
| Feb17-13, 07:15 AM | #3 |
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Recognitions:
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Isn't the integral of e^x w.r.t. x simply e^x + C?
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| Feb17-13, 10:54 PM | #4 |
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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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| calculus, change of variables, integratation |
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