Let $T>0$ and $a,b,x\in \mathbb{R}$ such that $a<b$.

Take $I:=\int_{-T}^T \frac{e^{-it(x-b)} - e^{-it(x-a)}}{-it} dt$.

Take $J:=2 \int_{x-b}^{x-a}\frac{\sin(Tu)}{u}du$.

One way to show $I=J$ is by using Fubini’s theorem.

In detail, $I=\int_{-T}^T \int_{x-b}^{x-a} e^{itu} du dt = \int_{x-b}^{x-a} \int_{-T}^T e^{itu} dt du = J$.

Is there another way to show that $I=J$? Are there sufficient “translation” and “linear isomorphism” whose composition can be used as change of variable?

up vote 1 down vote accepted

It's easy to see that $I(T=0)=J(T=0)=0$.

Take derivatives of both integrals with respect to $T$:

$ \frac{dI}{dT}=i\Big(\frac{e^{iT(x-b)}-e^{iT(x-a)}}{T}+\frac{e^{-iT(x-b)}-e^{-iT(x-a)}}{-T}\Big)=\frac{2}{T}\big(\sin(T(x-a))-\sin(T(x-b))\big)$

$\frac{dJ}{dT}=2\int\limits^{x-a}_{x-b}\cos(Tu)du=\frac{2}{T}\big(\sin(T(x-a))-\sin(T(x-b))\big)$

and thus we observe $$\frac{d(I-J)}{dT}=0\Leftrightarrow I-J=C, C\in\mathbb{R}$$.

But the initial condition imposes $C=0$ and the requested proof is complete.

P.S I modified the sign in the exponential on the first line to match the Fubini theorem derivation below and for the preceding proof to go through as indicated.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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