Author Topic: FE Sample--Problem 6  (Read 1454 times)

Victor Ivrii

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FE Sample--Problem 6
« on: November 27, 2018, 03:57:44 AM »
Calculate
$$
\int_0^\infty \frac{x\sin (x)}{1+x^4}.
$$

Hint:
Consider
$$
\int _\Gamma f(z)\,dz \qquad \text{with  } \ f(z)=\frac{ze^{iz}}{1+z^4}
$$
over contour $\Gamma$ on the picture below:
« Last Edit: November 27, 2018, 04:17:30 AM by Victor Ivrii »

Wanying Zhang

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Re: FE Sample--Problem 6
« Reply #1 on: November 27, 2018, 12:18:26 PM »
Here is the solution of problem 6.

Wanying Zhang

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Re: FE Sample--Problem 6
« Reply #2 on: November 27, 2018, 12:31:54 PM »
Sorry, I think I had a little computation mistake in the last photo.
Here is the new solution. Sorry for the inconvenience.

Victor Ivrii

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Re: FE Sample--Problem 6
« Reply #3 on: November 30, 2018, 04:20:15 AM »
Contour integral is calculated correctly, but then with integral
$$\int_{-\infty}^\infty \frac{xe^{ix}}{1+x^4}dx$$
you are wrong: $e^{-ix}$ appears only after you transform it
 $$\int_{-\infty}^\infty \frac{xe^{ix}}{1+x^4}dx= \int_0^\infty \frac{xe^{ix}}{1+x^4}dx +\int_{-\infty}^0 \frac{xe^{ix}}{1+x^4}dx =   \int_0^\infty \frac{xe^{ix}}{1+x^4}dx +\int_{\infty}^0 \frac{-xe^{-ix}}{1+x^4}dx =  \int_0^\infty \frac{x(e^{ix}-e^{-ix})}{1+x^4}dx = 2i I$$ .