Take the 2-minute tour ×
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.

I am studying for the exam and found the following formula on my notebook:

$\displaystyle \int_{r+1}^\infty (x-(r+1))g(x)dx$

=$\displaystyle \int_{r}^\infty (x-r)g(x)dx$

But I believe that this is not generally true unless $g$ is special such as a peridoic function.

I suspect that the above function is probably supposed to be

$\displaystyle \int_{r+1}^\infty (x-(r+1))g(x)dx$

=$\displaystyle \int_{r}^\infty (x-r)g(x+1)dx$

Could anyone help me confirm that?

share|improve this question
On the LHS integral, let $x\mapsto x+1$. Then you get what you claim. –  Pedro Tamaroff Mar 12 '13 at 4:12
Thank you for your answer! –  Tengu Mar 12 '13 at 4:23

2 Answers 2

up vote 1 down vote accepted

The second formula is correct in general. What you have in the first formula is wrong. You are correct.

share|improve this answer
Thank you very much! –  Tengu Mar 12 '13 at 4:24

That is correct. An easy spot check is that the integrand at the lower limit in both cases is $(0)(g(r+1))$ which the first one fails-it is $(0)(g(r+1))$ in the top and $(0)(g(r))$ in the second line. More formally you could do a $u$ substitution of $u=x-1$ and notice that (up to the name of the dummy variable) they match.

share|improve this answer
Thank you very much for your answer! –  Tengu Mar 12 '13 at 4:55

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.