# Calculate $\int_{- \infty}^{\infty} \frac{\sin \omega x}{x} dx$ for $\omega$ real and positive

Calculate $$\int_{- \infty}^{\infty} \frac{ \sin\omega x}{x} dx$$ for $\omega$ real and positive

The solution that's given in the textbook for my Complex Analysis course is the following:

Solution: Since $g(z) = \frac{\sin \omega z}{z}$ is analytic everywhere we may write

We then get

Now in the solution above, no explanation is given as to how we go from one equality to another in the first line. For example I'm not sure why the principal value of the integral of $\frac{\sin \omega x}{x}$ equals the principal value of the integral of the imaginary part of $$\frac{e^{i \omega x}}{x}$$ and I'm not sure of the reason why the next equality holds.

Could somebody please explain how we arrive at all of these equalities in the first line?

• Nothing changed after the first equality sign in your solution. And what does the line in the middle of the integral sign designate? – MrYouMath Jun 1 '18 at 15:50
• MrYouMath, it designates the principal value of the integral – Perturbative Jun 1 '18 at 15:51
• The fact that $e^{i\omega x}=\cos\omega x+i\sin\omega x$ has nothing to do with the integrals. If two expressions are equal then, of course, the integrals are equal too. – A.Γ. Jun 1 '18 at 15:54
• Is your question why $\sin\omega x=\mathrm{Im}\ e^{i\omega x}$? – CooperCape Jun 1 '18 at 15:55
• @CooperCape Yes and why the principal value of the integral of the imaginary part of $\frac{e^{i \omega x}}{x}$ equals the imaginary part of the principal value of the integral of $\frac{e^{i \omega x}}{x}$ – Perturbative Jun 1 '18 at 15:57

The middle equality $\sin\omega x=\operatorname{Im}e^{i\omega x}$ comes from Euler's formula.
The right equality comes from linearity of integral $$\int (g(x)+h(x))\,dx=\int g(x)\,dx+\int h(x)\,dx.$$ If you apply it to $f(x)=u(x)+iv(x)$ you'll get $$\int(u(x)+iv(x))\,dx=\int u(x)\,dx+i\int v(x)\,dx,$$ hence, the imaginary part of the integral is $\int v(x)\,dx=\int\operatorname{Im}f(x)\,dx$.