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#### chisigma

##### Well-known member

- Feb 13, 2012

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Integrale di analisi complessa • Forum Scienze Matematiche

…an Italian student failed to solve the following integral…

$\displaystyle\int_{0}^{\infty} \frac{\sin 2 t}{1+t^{3}}\ dt$

After several 'attempts' [that pratically meansthe use of Mathematica,WolframAlpha or other computer tools...] the student has been answered that the integral is 'too difficult'. Effectively if You instruct WolframAlpha to solve the integral...

$\displaystyle \int_{0}^{\infty} \frac{\sin 2t}{1+t^{n}}\ dt$ (2)

... the solution is possible for n=1 and n=2 but not for n>2. Because I think that is not a good idea to replace the brain with a computer, let me try to find a solution to integral (1).

The starting point is the following Laplace Transform relation...

$\displaystyle \mathcal{L} \{\frac{1}{t+a}\}=e^{a s}\ \text{Ei}\ (a s)$ (3)

... where Ei(*) if the Exponential Integral Function. The (3) permits us to arrive to a first nice result...

$\displaystyle \int_{0}^{\infty} \frac{e^{i\ \omega\ t}}{t+a}\ dt = e^{-i\ a\ \omega}\ \text{Ei} (-i\ a\ \omega)$ (4)

... that, remembering the identity involving Sine Integral, Cosine Integral and Exponential Integral functions...

$\displaystyle \text{Ei}\ (i x)= - \text{Ci} (x) + i\ [\frac{\pi}{2} + \text{Si} (x) ] $ (5)

… is equivalent to write...

$\displaystyle \int_{0}^{\infty} \frac{e^{i \omega\ t}}{t+a}\ dt = - \cos (a \omega)\ \text{Ci}(a \omega) + \sin (a \omega)\ [\frac{\pi}{2} - \text{Si} (a \omega)] +i\ \{\cos (a \omega)\ [\frac{\pi}{2} - \text{Si} (a \omega)] + \sin (a \omega)\ \text{Ci} (a \omega) \}$ (6)

… or alternatively…

$\displaystyle\int_{0}^{\infty} \frac{\cos \omega\ t}{t+a}\ dt = - \cos (a \omega)\ \text{Ci}(a\omega) + \sin (a \omega)\ [\frac{\pi}{2} - \text{Si} (a \omega)]$ (7)

$\displaystyle \int_{0}^{\infty} \frac{\sin\omega\ t}{t+a}\ dt = \cos (a \omega)\ [\frac{\pi}{2} - \text{Si} (a \omega)] + \sin (a \omega)\ \text{Ci} (a \omega)$ (8)

In a successive post we will see how to use these result in computation of (1)...

Kind regards

$\chi$ $\sigma$