# Gaussian Integral Substitution

1. Feb 17, 2013

### pcvrx560

$$\int_{-1}^{1}e^{x}dx$$

Then performing the substitution $x=\frac{1}{t}$ would give me

$$\int_{-1}^{1}-e^\frac{1}{t}t^{-2}dt$$

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.

Last edited: Feb 17, 2013
2. Feb 17, 2013

### tiny-tim

hi pcvrx560!
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 $-\int^{-1}_{-\infty} - \int^{\infty}_{1}$)​

3. Feb 17, 2013

### SteamKing

Staff Emeritus
Isn't the integral of e^x w.r.t. x simply e^x + C?

4. Feb 17, 2013

### pcvrx560

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.