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I have come across an integrl where they introduce a new variable $k' = k - k_0$ where $k_0 = \text{const}$.

$$ \begin{split} \psi(x) &= \int\limits_{-\infty}^{\infty} e^{-\frac{(k-k_0)^2}{2 \sigma_x^2}} e^{ikx} \,\,\mathrm{d}k\\ \boxed{\scriptsize k' = k-k_0} &\Longrightarrow \boxed{\scriptsize k=k' - k_0} \Longrightarrow\boxed{{\scriptsize \mathrm{d}k = \ ???}}\\ \psi(x) &= \int\limits_{-\infty}^{\infty} e^{- k'^2 / 2 \sigma_x^2} \, e^{i(k' -k_0)x} \,\,\mathrm{d}k'\\ \end{split} $$

I am nt sure how did an author get $\mathrm{d}k'$ out of $\mathrm{d}k$ so i need you to confirm my take on this. If i differentiate a new variable i get:

$$ \begin{split} \frac{d k'}{dk} &= \frac{d k}{d k} - \frac{d k_0}{dk}\\ \frac{d k'}{dk} &= \frac{d k}{d k}\\ \end{split} $$

Am i allowed to just cancel out the $dk$ in the denominator to get the below?

$$ dk' = dk $$

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Note that $\frac{dk}{dk}=1$ (we are not cancelling). As to whether you can replace $dk$ by $dk'$, the short answer is yes. And it is not really cancellation, but explaining what it is would take an answer, not a comment. –  André Nicolas Feb 26 '13 at 9:36
    
Oh i see so the proper way would be $\frac{dk'}{dk} = 1$ and therefore $dk' = dk$. I knew i was doing somthing wrong. Thank you. Please write an anwser so i can accept it. –  71GA Feb 26 '13 at 10:10
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1 Answer 1

up vote 1 down vote accepted

$\boxed{\scriptsize k=k' - k_0} \Longrightarrow\boxed{{\scriptsize \mathrm{d}k = \ ???}}$

When you have the first equation, derive each side by k, left side would be $\mathrm{d}k$, right side would be $\mathrm{d}k'$ and $k_0$ is just a constant so derivative would be zero. New equation would be:

$\boxed{\mathrm{d}k=\mathrm{d}k'}$

P.S. Sorry for bad language.

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