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?


1 Answer 1


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)$


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.


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