I am confused about if the text book set $z = x-y$, then why there is no $-$ after variable change? I think it is obvious that $dy = - dz$
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2 Answers
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They have compensated for this negative sign, but it's hidden in their $\int_R$ notation. Substitution affects both the $d$ term and the bounds of an integral, but we can't see that here. Writing it with explicit bounds, and inserting an intermediate calculation, we get $$ \int_{-\infty}^\infty [\cdots]dy=-\int_\infty^{-\infty}[\cdots]dz\\ =\int_{-\infty}^\infty [\cdots ]dz $$ Hopefully this is clearer.
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$\begingroup$ I did not get why substitution affects bound, could you please explain it for me? Thanks $\endgroup$– CooperNov 25, 2019 at 7:09
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$\begingroup$ @Cooper Look at any formula you have for substitution in a definite integral, in any book you may have. It will be there. Basically, the original integral was for when $y$ went from $-\infty$ to $\infty$. That corresponds to $z$ going from $\infty$ to $-\infty$. $\endgroup$– ArthurNov 25, 2019 at 7:13
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You are right there is a minus sign but also the boundaries of the integral will change after the change of variables, so the integral will be
$-\int_{+\infty}^{-\infty}=\int_{-\infty}^{+\infty}= \int_{\mathbb R}$
answered Nov 25, 2019 at 6:24
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$\begingroup$ so integrate from negative infinity to positive infinity is the same as to integrate it from positive infinity to negative infinity? $\endgroup$– CooperNov 25, 2019 at 6:28
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$\begingroup$ No it is the opposite of it, $\int_a^b=-\int_b^a$ $\endgroup$ Nov 25, 2019 at 6:30