I want to evaluate the integral:
∫0−∞2x2−1x4+1dx
using contour integration.
I re-wrote it as: ∫∞02x2−1x4+1dx. I am considering of integrating on a semicircle contour with center at the origin. I considered the function f(z)=2z2−1z4+1 which has 4 simple poles but only two of them lie on the upper half plane and included in the contour which are: z1=1+i√2,z2=−1+i√2.
The residue at z1 equals Res(f;z1)=−2i−12√2 while the residue at z2 equals −2√2i−2√2. (if I have done the calculations right)
Now, I don't know how to continue. Should I find the residues at the other poles as well and the say ∮Cf(z)=2πi∑res where C is the semicircle contour and then expand it? That is:
∮Cf(z)dz=∫a0+∫arc
Then let a→+∞ then than arc integral would go to zero.
But I don't know how to proceed.
I had dealt with this integral with residues converting it into a minus infinity to infinity integral but with contours I am having a bit of problem.
Therefore I'd like some help.
Answer
You could also use the following contour.
∫0−∞2x2−1x4+1dx=∫+∞02x2−1x4+1dx
Has an analytic continuation as ∫Γ2z2−1z4+1dz with 4 poles, but just on pole inside of the contour.
res
I think you made a calculation error here (?)
Using \int_{\Gamma} \frac{2z^2-1}{z^4+1}\operatorname dz = \color{blue}{\int_{\Gamma_1}\frac{2z^2-1}{z^4+1}\operatorname dz} + \int_{\Gamma_2}\frac{2z^2-1}{z^4+1}\operatorname dz + {\color{red}{\int_{\Gamma_3}\frac{2z^2-1}{z^4+1}\operatorname dz}}
- Now \color{blue}{\int_{\Gamma_1} \to 0} as R\to +\infty which can be proven using the triangle inequality.
- Use \Gamma_2 \leftrightarrow z(x) = x and x:0\to R
- Use \color{red}{\Gamma_3 \leftrightarrow z(y) = iy} and y: R \to 0
Which finally results in (as R \to +\infty)
2\pi i \left(\frac{3\sqrt{2}}{8} -i \frac{\sqrt{2}}{8}\right) = \color{blue}{0}+\int_{0}^{+\infty}\frac{2x^2-1}{x^4+1}\operatorname dx + \color{red}{ i \int_{+\infty}^0\frac{-2y^2-1}{y^4+1}\operatorname dy}
Here you can read the real parts which results in:
\int_{0}^{+\infty}\frac{2x^2-1}{x^4+1}\operatorname dx = \frac{\pi\sqrt{2}}{4}
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