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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!

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  • $\begingroup$ 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
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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!

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  • $\begingroup$ 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

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