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Before I proceed here's a quick disclaimer: English is not my first language, some of the terminology I used might be incorrect.

I tried solving it and my final solution was not a real number (which would infer I did something fundamentally wrong).

$I=\frac{1}{2} \int_{-\infty}^\infty \frac{z\sin 3z}{z^2+1} dz$

I tried finding the roots of $z^2+1$ which ended up being $\pm i$

Knowing only i is part of the contour in the upper-half plane, that would mean $Rez(f, -i) = 0$ and therefore I would only need to calculate $Rez(f, i)$

$z=i$ seemed to be a first degree pole, therefore $Rez(f, i) = \frac{1}{0!} \displaystyle\lim_{z\rightarrow i} ((z-i) f(z)) = \lim_{z\rightarrow i} (z-i)\frac{z\sin 3z}{(z-i)(z+i)} = \frac{i \sin 3i}{2i} = \frac{\sin 3i}{2} = \frac{e^{-3} -e + 3}{2}$

That seems all well and good until I realized that the integral is now $I = i \pi \frac{e^{-3} -e + 3}{2}$ which would leave me with a complex solution.

Where did I go wrong?

Oreoezi
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  • You must use $\sin(3z)=\text{Im}(e^{i3z})$. – OnTheWay Jun 07 '23 at 11:14
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    See https://math.stackexchange.com/q/632530/42969 for a very similar question. – Martin R Jun 07 '23 at 11:14
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    First of all, always replace trigonometric functions by exponentials when you define the complex function. You'll not be able to bound $\sin(3z)$ on the semicircle which is part of your contour. So replace it by $e^{3iz}$, and at the end just take the imaginary part. – Mark Jun 07 '23 at 11:17
  • Conceivably, it is sometimes useful to use $\sin$ as just $\sin$. But in this case the error term - the integral around the semicircle - diverges if you use $\sin$, but vanishes if you use the (correct) exponential – FShrike Jun 07 '23 at 11:20
  • $|\frac {e^{3iz} - e^{-3iz}}{2i}|$ goes off to infinity as $\Im(z)$ approaches infinity. The integral does not converge along the contour. For this reason, we need a different formula for $\sin 3z$ – user317176 Jun 07 '23 at 11:34

2 Answers2

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Take $$f(z)=\frac{ze^{3iz}}{z^2+1},$$ now $$\operatorname{Res}(f,i)=\frac{ie^{-3}}{2i}=\frac{e^{-3}}{2},$$ so the integral is $\pi ie^{-3}.$ Now since $e^{3iz}=\cos (3z)+i\sin (3z),$ the integral desired is $\operatorname{Im}(i\pi e^{-3})=\pi e^{-3},$ so $$\int_0^\infty \frac{x\sin (3x)}{x^2+1}dx=\frac{\pi e^{-3}}{2}.$$

Guillermo García Sáez
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Do not abuse a theorem; see comments for more information.

For the sake of clarity, let me quote a theorem taken from a book (in Chinese):

Suppose that $P(z)$ and $Q(z)$ are real polynomials that are relatively prime. Suppose that the degree of $Q(z)$ is higher than that of $P(z)$. Suppose that $Q(x) \neq 0$ for any real number $x$. Suppose that $m$ is a positive integer. Then $$ \begin{aligned} & \int_{-\infty}^{\infty} {\frac{P(x)}{Q(x)} \mathrm{e}^{\mathrm{i}mx}\,\mathrm{d}x} \\ = {} & 2\pi \mathrm{i} \cdot (\text{the sum of the residues of the poles of $\dfrac{P(z)}{Q(z)}\mathrm{e}^{\mathrm{i}mx}$ with positive imaginary part}). \end{aligned} $$

Put $P(z) = z$, $Q(z) = z^2 + 1$ and $m = 3$. The only pole with positive imaginary part of $\dfrac{P(z)}{Q(z)}\mathrm{e}^{\mathrm{i}mx}$ is $\mathrm{i}$. It is not hard to see that the residue is $\dfrac{\mathrm{e}^{-3}}{2} = \dfrac{1}{2\mathrm{e}^3}$. Hence $$ \int_{-\infty}^{\infty} {\frac{x}{x^2+1} \mathrm{e}^{\mathrm{i}3x}\,\mathrm{d}x} = \mathrm{i} \frac{2\pi}{2\mathrm{e}^3}. $$ It is not hard to see that $$ \int_{-\infty}^{\infty} {\frac{x}{x^2+1} \sin {3x} \,\mathrm{d}x} = \frac{2\pi}{2\mathrm{e}^3}. $$ Hence $$ \int_{0}^{\infty} {\frac{x}{x^2+1} \sin {3x} \,\mathrm{d}x} = \frac{2\pi}{4\mathrm{e}^3}. $$

Juliamisto
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