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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?

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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.

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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.

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

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