# How does $\int_{z=-R+0i}^{R+0i} \frac{e^{2iz}-1-2iz}{z^2}\ dx$ become $\int_{-R}^R \frac{\sin^2x}{x^2}\ dx$?

While trying to compute $\int_0^\infty \frac{\sin^2 x}{x^2}\ dx$, the author of this book suggests computing $\int_{C_R} \frac{e^{2iz}-1-2iz}{z^2}\ dz$ on a semi-circular contour in the upper half-plane.

The singularity at $z=0$ is removable, so the function is entire, so the integral becomes zero. He then makes a huge jump and says, "Thus, $-2\int_{-R}^R \frac{\sin^2 x}{x^2}\ dx -2i\int_{\Gamma_R} \frac{dz}{z} + \int_{\Gamma_R} \frac{e^{2i z}-1}{z^2}\ dz = 0$."

Here, $\Gamma_R$ denote the "arc" part of the semi-circular contour.

I get where he gets the second terms from. I don't get where he gets the first.

How do we go from

$$\int_{-R}^R \frac{e^{2ix}-1-2ix}{x^2}\ dx$$

to

$$\int_{-R}^R \frac{\sin^2 x}{x^2}\ dx?$$

• I'm referring to the integral $\int_{C_R} \frac{e^{2iz}-1-2iz}{z^2}\ dz$. – Emily Mar 19 '13 at 23:14

$$e^{2iz}-1-2iz=1+2iz+\frac {(2iz)^2}2+\frac{(2iz)^3}6+\ldots\frac {(2iz)^n}{n!}+\ldots-1-2iz=\\ =\frac{(2iz)^2}2+\frac{(2iz)^3}{3!}+\ldots+\frac{(2iz)^{n+2}}{(n+2)!}+\ldots$$ If you integrate it over $(-R, R)$, obviously all odd powers will drop out since their antiderivatives will be even. So let's consider even powers only $$e^{2iz}-1-2iz \stackrel{\int}{\equiv} \frac {(2ix)^2}{2!} + \frac {(2ix)^4}{4!} + \frac {(2ix)^6}{6!} + \ldots + \frac {(2ix)^{2k+2}}{(2k+2)!} + \ldots = \\ = -\frac {(2x)^2}{2!} + \frac {(2x)^4}{4!} - \frac {(2x)^6}{6!} + \ldots + (-1)^{k+1} \frac {(2x)^{2k+2}}{(2k+2)!} + \ldots = \\ = 1 -\frac {(2x)^2}{2!} + \frac {(2x)^4}{4!} - \frac {(2x)^6}{6!} + \ldots + (-1)^{k+1} \frac {(2x)^{2k+2}}{(2k+2)!} + \ldots - 1 = \\ = \cos 2x-1 = -2\sin^2x$$

An idea:

$$\frac{e^{2ix}-1-2ix}{x^2}=\frac{\cos 2x+i\sin 2x-1-2ix}{x^2}=\frac{\rlap{/}1-2\sin^2x+i\sin 2x-\rlap{/}1-2ix}{x^2}=$$

$$=-2\frac{\sin^2 x}{x^2}+\frac{\sin 2x-2x}{x^2}i$$

and now perhaps that author (what book is that, BTW?) meant to take the real part of that complex function integral...?

• I think it should be $\sin2x -2x$ – Kaster Mar 20 '13 at 1:15
• Also, integral of the imaginary part is simply zero, so the whole integral is equal to real part anyway. – Kaster Mar 20 '13 at 1:20
• Don, this is Complex Analysis by Joseph Bak and Donald Newman. – Emily Mar 20 '13 at 1:51
• Yes to both your comments, @Kaster. Thanks – DonAntonio Mar 20 '13 at 2:15
• Yes @Arkamis, I have it and already checked it. It seems to work fine, taking into account that $\,\frac{\sin^2x}{x^2}\,$ is an even function... – DonAntonio Mar 20 '13 at 2:20