Line Integrals in Wolfram Alpha?

Is it possible to calculate a line integral in wolfram alpha, and if so what would be the form I would have to put it in?

Answer 1

Hello

Here is an example:

http://www.wolframalpha.com/input/?i=%E2%88%AB%28x…

I hope this is what you mean by “line integral”)

Regards

Answer 2

Line Integral Calculator

Answer 3

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RE:
Line Integrals in Wolfram Alpha?
Is it possible to calculate a line integral in wolfram alpha, and if so what would be the form I would have to put it in?

Answer 4

line integrals wolfram alpha

Answer 5

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MathPhD is right (as, with a handle such as that, we would hope): the exact computation of this integral is a very neat application of complex variables to real-valued functions. This is covered as a standard application in a first course on complex analysis, but it really is quite fascinating. [1] describes this process in more detail than I care to, but in a nutshell, the idea goes like this: * Treat the integrand as a function of a complex variable, and note that, on the real axis, cos z/[(z^2 + 1)(z^2 + 2)] = Re e^(iz)/[(z^2 + 1)(z^2 + 2)] = Re f(z) * Consider the contour integral of f(z) around a semicircle, centered at the origin and oriented counter-clockwise, of radius R (if you’ve studied any multivariable calculus, this is analogous to a path or line integral in two real dimensions). * Use your complex variables theorem of choice (a calculation of residues [2], together with the Cauchy residue theorem [3], will make short work of the contour integral of f around the semicircle C_R). * Notice that this integral can be decomposed into two parts: an integral over a semicircular arc with nonzero imaginary part traced from (R, 0) to (-R, 0) parameterized by (R cos t, R sin t) on (0, pi), and an integral over a real line segment from -R to R. The integral over the real segment is the same as the real integral in the context you’re used to. * Consider the limit of this integral as R –> infinity. We can show that, since e^(iz) is bounded in magnitude by 1, that the contour integral over the circular part goes to 0 as R –> inf. Hence, the complex contour integral derives its value entirely from the integral over the real axis. Since the integral is only nonzero on that axis, take its real part to obtain the value of the desired integral.

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