Through the course of a problem I am working on I have reached two integrals that look similar to some of the trigonometric integrals. The integrals I have are the following:

$$\int_{0}^{\infty} \frac{\sin(x+i)}{x+ i } \, \mathscr{d}x$$ and $$\int_{0}^{\infty} \frac{\cos(x+i)}{x+ i } \, \mathscr{d}x$$

I do not know much about complex number theory, and am nervous to evaluate these integrals without help. However, I have evaluated these integrals in Maple and got that

$$\int_{0}^{\infty} \frac{\sin(x+i)}{x+ i } \, \mathscr{d}x = - \int_{i}^{\infty} \frac{\sin(t)}{t } \, \mathscr{d}t = -\text{Si}(i) + \frac{\pi}{2} $$ and $$\int_{0}^{\infty} \frac{\cos(x+i)}{x+ i } \, \mathscr{d}x =- \int_{0}^{i} \frac{\cos(t)}{t} \, \mathscr{d}t = -\text{Ci}(i)$$

All I am looking is for some explanation (that is suitable for someone with very little experience with complex numbers) as to how Maple obtained this solution. Thanks!

  • $\begingroup$ is $i = \sqrt{-1}$ or some parameter? $\endgroup$ – Gregory Nov 3 '15 at 0:42
  • $\begingroup$ sorry, yes $i=\sqrt{-1}$. If this is not the proper way to put this into TeX feel free to edit! $\endgroup$ – möbius Nov 3 '15 at 0:43
  • $\begingroup$ If you take it as given that $\int_0^\infty \sin x/x \, dx = \pi/2$, Then change of variables $t = x + i$ gives maples result. To see three easy ways to evaluate the integral, see here $\endgroup$ – Gregory Nov 3 '15 at 0:52
  • $\begingroup$ @Gregory Yeah I was browsing the Dirichlet integral earlier, but have never heard of it before today and was not sure whether it was valid for complex numbers. Anyway, I would still need a separate method to evaluate the cosine integral. In addition, I was not sure how to do the change of variable for complex numbers; i.e. after the change, is the upper limit of the integral now $\infty + i$? I was not sure how this works $\endgroup$ – möbius Nov 3 '15 at 0:56
  • $\begingroup$ @mobius After making the substitution $x+i \to x$, you can deform the contour by using Cauchy's Integral Theorem. $\endgroup$ – Mark Viola Nov 3 '15 at 1:22

Let $I$ be the integral given by

$$I=\int_0^\infty \frac{\sin (x+i)}{x+i}\,dx \tag 1$$

We will use Cauchy's Integral Theorem to evaluate the integral in $(1)$.

Let $f$ be the function given by $f(z)=\frac{\sin z}{z}$ and let $\gamma$ be the closed contour comprised of the four line segments

(i) from $(0,0)$ to $(R,0)$;

(ii) from $(R,0)$ to $(R,1)$;

(iii) from $(R,1)$ to $(0,1)$;

(iv) from $(0,1)$ to $(0,0)$.

Since $f$ is analytic within the region bounded by $\gamma$, we have from Cauchy's Integral Theorem

$$\oint_\gamma \frac{\sin z}{z}\,dz=0 \tag 2$$

We can also write the closed contour integral in $(2)$ as

$$\begin{align} \oint_\gamma \frac{\sin z}{z}&=\int_0^R \frac{\sin x}{x}\,dx+\int_0^1\frac{\sin (R+iy)}{R+iy}\,idy\\\\ &+\int_{R+i}^i \frac{\sin x}{x}\,dx+\int_{i}^0 \frac{\sin z}{z}\,dz \tag 3 \end{align}$$

As $R\to \infty$, the first integral on the right-hand side of $(3)$ approaches the Dirichlet Integral, which has a value of $\pi/2$.

The second integral goes to $0$ as $R\to \infty$ since

$$\left|\frac{\sin (R+iy)}{R+iy}\right|=\sqrt{\frac{\sin^2(R)+\sinh^2(y)}{R^2+1}} \le \frac{\sqrt{1+e^2/4}}{R}$$

The third integral goes to $-1$ times the integral of interest $\int_i^{\infty+i}\frac{\sin x}{x}\,dx$.

And finally, the fourth integral is $-1$ times the Sine Integral, $\text{Si}(x)=\int_0^x\frac{\sin z}{z}\,dz$ evaluated at $x=i$.

Putting everything together, we can write, therefore,

$$\int_0^\infty \frac{\sin(x+i)}{x+i}\,dx=\frac{\pi}{2}-\text{Si}(i)$$

as expected!

  • $\begingroup$ Excellent! Very clear, thank you for taking the time to do this! $\endgroup$ – möbius Nov 3 '15 at 11:47
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Nov 3 '15 at 15:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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