Practically, derivatives always have some "plug and chug" algorithm to get them. In applications, the value of C would be determined by the initial conditions of some differential equation.Įven for the most pathological functions, the derivative is defined by a limit and can either be evaluated or shown not to exist. This is because the derivative of any constant is 0. Actually, if an antiderivative can be found, then actually an infinite family of functions will do the trick: any non-constant function f(x) + C, an arbitrary constant. Another antiderivative of 1/x 5 is -1/( 4x 4 ) + 69. For example, an antiderivative of 1/x 5 is -1/( 4x 4 ) + 6. When you find an antiderivative of a function, you look for a new function such that, when you take its derivative, you get back the original function. I get it, haha funny proof left as exercise to the reader, never heard that one before amirite? But it really is cringy to me, and probably many others having completed complex analysis, as the first step just does not make any sense.īoth expressions are called "indefinite integrals" or more accurately "antiderivatives". You get the point, it would not start by letting y = x 5 + 1. By the Residue Theorem from complex analysis, we have that the integral of 1/(x 5 + 1) on C_R w.r.t. Something to start off with along the line of an actual calculation of the integral is the following: for R > 1, consider the counter-clockwise contour C_R formed by the interval together with the linesegment from 0 to Rt, with t = e 2pi*i/5, glued together via the shorter path of the circle with radius R and center 0. If you want to calculate the definite integral from 0 to infinity (which I guess the original responder was referring to), it is very useful to consider complex analysis, and the response starting with y = x 5 + 1 is just not the way to go. It's a fun geometry/trig exercise to find it using a clever isosceles triangle or some trig double/triple angle formulas.ĭude, really? Besides that what you are writing is just plain wrong, the original response was about complex analysis. This lets you write everything down more "numerically", which is why the solution Wolfram shows radicals inside the ln and arctan functions instead of sin and cos of multiples of 2pi/5. Incidentally, there is also a geometry trick to rewrite the coordinates as radicals - sort of like the "exact values" tricks you learn when finding the sin and cos of pi/6, pi/4, pi/3, etc. Having all the complex roots lets you factor the nasty quartic over R - just multiply the factors (x-z)(x-z*) corresponding to complex conjugate solutions to get the two irreducible quadratics that the quartic factors into. Alternatively, it's a well-known fact from field theory that the nth roots of -1 sit at the vertices of a regular n-gon in the complex plane, so you can use trig to find their coordinates. There's the obvious root x=-1, and so you can use long division to factor out x+1, but then you get a nasty quartic with no obvious factorization strategy. You don't have to, but I don't know of an obvious way of factoring x^5+1 over R.
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