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I have an integral where I need to change variables. The integral has the form,

$\int_0^x f(x,t) dt$ .

I change variables/rescale by setting $\tilde{t}=xt$, which means $d\tilde{t}=xdt$. Would the new integral have the following form & bounds:

$\int_0^{x^2} f(x,\tilde{t}) d\tilde{t}$ ?

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  • $\begingroup$ Why did you substitute $t$ with $\tilde{t}$ directly, clearly $t = \tilde{t}/x$. $\endgroup$
    – lightxbulb
    Feb 8, 2019 at 1:25
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    $\begingroup$ You have $\int_0^x f(x, t)dt$ and you want to make the change of variable $s= xt$. Yes, $ds= x dt$ so that $dt= \frac{1}{x}ds$. When t= 0, s= 0 and wen t= x, $s= x^2$. The integral is $\frac{1}{x}\int_0^{x^2} f(x, s/x)ds$. $\endgroup$
    – user247327
    Feb 8, 2019 at 1:26

1 Answer 1

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As you correctly computed, $d\widetilde{t}=xdt$, so $dt=d\widetilde{t}/x$. We can see by $\widetilde{t}=xt$ that $0\le t\le x$ means that $0\le \widetilde{t}\le x^2$ using $\widetilde{t}/x=t$. Plugging this all in, we get $$ \int_0^x f(x,t)dt=\int_0^{x^2} \frac{f(x,\widetilde{t}/x)}{x}d\widetilde{t}.$$

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  • $\begingroup$ What about the argument of $f$? $\endgroup$
    – lightxbulb
    Feb 8, 2019 at 1:31
  • $\begingroup$ You're right, of course. $\endgroup$ Feb 8, 2019 at 1:32

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