- #1

- 9

- 0

[itex]\int^{\infty}_{-\infty}\frac{e^{ix}dx}{\sqrt{a - x^2 + ib}}[/itex]

where a and b are real constants. If you have any suggestions I would be more than grateful.

Thanks in advance.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter freechus9
- Start date

- #1

- 9

- 0

[itex]\int^{\infty}_{-\infty}\frac{e^{ix}dx}{\sqrt{a - x^2 + ib}}[/itex]

where a and b are real constants. If you have any suggestions I would be more than grateful.

Thanks in advance.

- #2

- 336

- 0

- #3

- 1,800

- 53

- #4

- 9

- 0

For example, along the cut on the right side, x would have the value re[itex]^{i*\pi/8}[/itex]. Therefore x[itex]^{2}[/itex] is r[itex]^{2}[/itex]e[itex]^{i*\pi/4}[/itex]?

Along the cut on the left side, x would have the value re[itex]^{-i*15\pi/8}[/itex]. Therefore x[itex]^{2}[/itex] is r[itex]^{2}[/itex]e[itex]^{-i*15\pi/4}[/itex]?

Is this correct?

- #5

- 1,800

- 53

The value on one side of a square-root branch-cut is just the negative of the other side. Remember, the multifunctioned square root is [itex]\pm [/itex]. So on the lower side, the integral would be:

[tex]\int_{\infty}^{2^{1/4}} \frac{e^{iz}}{\sqrt{z_0-z^2}} e^{\pi i/8}dr,\quad z\to re^{\pi i/8}[/tex]

and on the other side, it would be:

[tex]\int_{2^{1/4}}^{\infty} \frac{e^{iz}}{-\sqrt{z_0-z^2}} e^{\pi i/8}dr,\quad z\to re^{\pi i/8}[/tex]

and after analyzing all the other parts of the contour and simplifying, we can write:

[tex]\int_{-\infty}^{\infty} \frac{e^{iz}}{\sqrt{z_0-z^2}}dz=2\int_{\infty}^{2^{1/4}} \frac{e^{iz}}{-\sqrt{z_0-z^2}} e^{\pi i/8}dr,\quad z\to re^{\pi i/8}[/tex]

at least for this problem. May have other issues if z0 is on the real axis or may have to modify the analysis somewhat.

Not sure if that's any easier to evaluate analytically than the original problem. May be or probably is an easier way to evaluate it but it's good practice to do it any way possible to help understand and perhaps lead to a better way of doing it.

[tex]\int_{\infty}^{2^{1/4}} \frac{e^{iz}}{\sqrt{z_0-z^2}} e^{\pi i/8}dr,\quad z\to re^{\pi i/8}[/tex]

and on the other side, it would be:

[tex]\int_{2^{1/4}}^{\infty} \frac{e^{iz}}{-\sqrt{z_0-z^2}} e^{\pi i/8}dr,\quad z\to re^{\pi i/8}[/tex]

and after analyzing all the other parts of the contour and simplifying, we can write:

[tex]\int_{-\infty}^{\infty} \frac{e^{iz}}{\sqrt{z_0-z^2}}dz=2\int_{\infty}^{2^{1/4}} \frac{e^{iz}}{-\sqrt{z_0-z^2}} e^{\pi i/8}dr,\quad z\to re^{\pi i/8}[/tex]

at least for this problem. May have other issues if z0 is on the real axis or may have to modify the analysis somewhat.

Not sure if that's any easier to evaluate analytically than the original problem. May be or probably is an easier way to evaluate it but it's good practice to do it any way possible to help understand and perhaps lead to a better way of doing it.

Last edited:

- #6

- 9

- 0

Thank you very much!

- #7

- 1,800

- 53

and after analyzing all the other parts of the contour and simplifying, we can write:

[tex]\int_{-\infty}^{\infty} \frac{e^{iz}}{\sqrt{z_0-z^2}}dz=2\int_{\infty}^{2^{1/4}} \frac{e^{iz}}{-\sqrt{z_0-z^2}} e^{\pi i/8}dr,\quad z\to re^{\pi i/8}[/tex]

That's not correct. Remember it's the sum over all the contours and by Cauchy's Theorem, that's zero so I should have written:

[tex]\int_{-\infty}^{\infty} \frac{e^{iz}}{\sqrt{z_0-z^2}}dz+\left(2\int_{\infty}^{2^{1/4}} \frac{e^{iz}}{-\sqrt{z_0-z^2}} e^{\pi i/8}dr,\quad z\to re^{\pi i/8}\right)=0[/tex]

so that:

[tex]\int_{-\infty}^{\infty} \frac{e^{iz}}{\sqrt{z_0-z^2}}dz=\left(-2\int_{\infty}^{2^{1/4}} \frac{e^{iz}}{-\sqrt{z_0-z^2}} e^{\pi i/8}dr,\quad z\to re^{\pi i/8}\right)[/tex]

However when I compute them numerically in Mathematica, there is a sign difference. I suspect that's because Mathematica is integrating over two different branches but I'm not sure.

Share: