Cauchy Principal Value calculation
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I am self-studying the residue theorem and its applications and I tried solving a problem which involves finding the principal value for an improper integral but I am not sure if my approach/answer is correct.
I wish to find $P.V. int_{-infty}^{infty} frac{dx}{x(x^2+1)}$
I know that the integrand has a singularity at x = 0 on the real line, and this singularity is a simple pole since $frac{d}{dx} x(x^2 + 1) = 3x^2 + 1$ which is not equal to zero at z = 0
I also know that the integrand has a simple pole at $z = i$ on the upper half plane and so I should have that
$$P.V. int_{-infty}^{infty} frac{dx}{x(x^2+1)} = 2 pi i Res(f, i) + pi i Res(f,0) = 2 pi i (frac{1}{-2}) + pi i = 0$$
Does this look right, the answer being zero concerns me for some reason but I assume it seems true since the function may be symmetric
complex-analysis contour-integration cauchy-principal-value
asked Dec 12 '18 at 19:05
Richard VillalobosRichard Villalobos
1757
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add a comment |
$begingroup$
I am self-studying the residue theorem and its applications and I tried solving a problem which involves finding the principal value for an improper integral but I am not sure if my approach/answer is correct.
I wish to find $P.V. int_{-infty}^{infty} frac{dx}{x(x^2+1)}$
I know that the integrand has a singularity at x = 0 on the real line, and this singularity is a simple pole since $frac{d}{dx} x(x^2 + 1) = 3x^2 + 1$ which is not equal to zero at z = 0
I also know that the integrand has a simple pole at $z = i$ on the upper half plane and so I should have that
$$P.V. int_{-infty}^{infty} frac{dx}{x(x^2+1)} = 2 pi i Res(f, i) + pi i Res(f,0) = 2 pi i (frac{1}{-2}) + pi i = 0$$
Does this look right, the answer being zero concerns me for some reason but I assume it seems true since the function may be symmetric
complex-analysis contour-integration cauchy-principal-value
asked Dec 12 '18 at 19:05
Richard VillalobosRichard Villalobos
1757
$endgroup$
3
$begingroup$
The integrand is odd.
$endgroup$
– gammatester
Dec 12 '18 at 19:09
add a comment |
$begingroup$
I am self-studying the residue theorem and its applications and I tried solving a problem which involves finding the principal value for an improper integral but I am not sure if my approach/answer is correct.
I wish to find $P.V. int_{-infty}^{infty} frac{dx}{x(x^2+1)}$
I know that the integrand has a singularity at x = 0 on the real line, and this singularity is a simple pole since $frac{d}{dx} x(x^2 + 1) = 3x^2 + 1$ which is not equal to zero at z = 0
I also know that the integrand has a simple pole at $z = i$ on the upper half plane and so I should have that
$$P.V. int_{-infty}^{infty} frac{dx}{x(x^2+1)} = 2 pi i Res(f, i) + pi i Res(f,0) = 2 pi i (frac{1}{-2}) + pi i = 0$$
Does this look right, the answer being zero concerns me for some reason but I assume it seems true since the function may be symmetric
complex-analysis contour-integration cauchy-principal-value
asked Dec 12 '18 at 19:05
Richard VillalobosRichard Villalobos
1757
$endgroup$
I am self-studying the residue theorem and its applications and I tried solving a problem which involves finding the principal value for an improper integral but I am not sure if my approach/answer is correct.
I wish to find $P.V. int_{-infty}^{infty} frac{dx}{x(x^2+1)}$
I know that the integrand has a singularity at x = 0 on the real line, and this singularity is a simple pole since $frac{d}{dx} x(x^2 + 1) = 3x^2 + 1$ which is not equal to zero at z = 0
I also know that the integrand has a simple pole at $z = i$ on the upper half plane and so I should have that
$$P.V. int_{-infty}^{infty} frac{dx}{x(x^2+1)} = 2 pi i Res(f, i) + pi i Res(f,0) = 2 pi i (frac{1}{-2}) + pi i = 0$$
Does this look right, the answer being zero concerns me for some reason but I assume it seems true since the function may be symmetric
complex-analysis contour-integration cauchy-principal-value
complex-analysis contour-integration cauchy-principal-value
asked Dec 12 '18 at 19:05
Richard VillalobosRichard Villalobos
1757
asked Dec 12 '18 at 19:05
Richard VillalobosRichard Villalobos
1757
asked Dec 12 '18 at 19:05
Richard VillalobosRichard Villalobos
1757
asked Dec 12 '18 at 19:05
Richard VillalobosRichard Villalobos
1757
asked Dec 12 '18 at 19:05
Richard VillalobosRichard Villalobos
1757
1757
3
$begingroup$
The integrand is odd.
$endgroup$
– gammatester
Dec 12 '18 at 19:09
add a comment |
3
$begingroup$
The integrand is odd.
$endgroup$
– gammatester
Dec 12 '18 at 19:09
3
3
$begingroup$
The integrand is odd.
$endgroup$
– gammatester
Dec 12 '18 at 19:09
$begingroup$
The integrand is odd.
$endgroup$
– gammatester
Dec 12 '18 at 19:09
add a comment |
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$begingroup$
The integrand is odd.
$endgroup$
– gammatester
Dec 12 '18 at 19:09