Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 1 down vote favorite

I got $dx=-t^{-2}dt$, giving me

$$ \int_{0}^{1}-\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2t^{2}}}t^{-2}dt $$

I can already tell that the above equation can't be right since it is negative everywhere. It actually evaluates to 1 minus the actual answer.

What am I doing wrong and what would be the correct limits?

EDIT: What if my integral were

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

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?

share|improve this question
1  
Your change of variables has a singularity when $x=0$. – Antonio Vargas Feb 17 at 7:51
$x=1/t$ and $-\infty\lt x\lt1$ does not yield $0\lt t\lt1$. – Did Feb 17 at 7:59
Why can't I set the lower limit to be $a$ and let $a\rightarrow -\infty$? – G.P. Burdell Feb 17 at 8:02
Good idea. Try to see what happens with $-42\lt x\lt1$ when $x=1/t$. Where will be $t$? – Did Feb 17 at 8:04
1  
$\int_{-\infty}^{1} f(x) dx = \left[\int_{-\infty}^{0-} + \int_{0+}^{1}\right] f(x) dx = \left[\int_{0}^{-\infty} + \int_{\infty}^{1}\right] f(\frac{1}{t}) (-\frac{dt}{t^2} )$ – achille hui Feb 17 at 8:16
show 3 more comments

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.