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## Main Question or Discussion Point

if we apply ordinary integration formula to it we'll get [t^0 / 0] regardless of the limits of integration. can someone show me how to prove mathematically that this method is invalid to use and why

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if we apply ordinary integration formula to it we'll get [t^0 / 0] regardless of the limits of integration. can someone show me how to prove mathematically that this method is invalid to use and why

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mathwonk

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so you just have no reason to use that rule on this integral.

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why did they exclude -1?

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radou

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Your question is equivalent to the question 'Can we divide by zero?'.why did they exclude -1?

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HallsofIvy

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What do you mean by "prove mathematically". t^0/0= 1/0 is "undefined". If we set 1/0= x then we have 1= 0*x which is not true for any possible x.if we apply ordinary integration formula to it we'll get [t^0 / 0] regardless of the limits of integration. can someone show me how to prove mathematically that this method is invalid to use and why

That's why every text book gives the "power rule" for integration with the proviso "n not equal to -1".

Many books define a "new function" by

[tex]log(x)= \int_0^x \frac{1}{t} dt[/tex]

and then prove that this is, in fact, the usual natural logarithm function.

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help would be much appreciated as this is fundamental in calculus 2 :D pls clear it up 4 me

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HallsofIvy

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Yes, it is still true that 1/t has an anti-derivative (everywhere except at 0). Since it is continuous for t not equal to 0, it must have an anti-derivative.

The fact that 1/0 us UNDEFINED simply means that the formula "anti-derivative of [itex]x^n[/itex] is [itex]\frac{1}{n+1}x^{n+1} + C[/itex]" does not apply when n= -1. That doesn't mean that some other formula will not apply.

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mathwonk

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take the derivative, geting n.t^(n-1).

can this ever equal 1/t? tht would require n =m 0, but then the zero out fron would still not let this equal 1/t.

so you never get 1/t this way.

i.e. the function 1/t is never obtained by differentiating using the power rule.

so sad.

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mathwonk

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well its a bit abstract. you have to prove somehow that the natural logarithm function equals the area function for1/x, so you need some way to recognize a log function even in disguise.

so basically you prove that any non constant continuous function L from R+ to R, which saisifies the law L(xy) = Lx)+L(y), must be a log fucntion, and the base of the logs is the unique number a such that L(a) = 1.

then you prove that since the area function for 1/x (taken from 1 to x) has derivative 1/x, it lso satisifes those laws. hence it must be a log function!

and by approximating integrals you can show there is some number e between 2 and 3 such that L(e) = 1. and that e is the base.

so basically you prove that any non constant continuous function L from R+ to R, which saisifies the law L(xy) = Lx)+L(y), must be a log fucntion, and the base of the logs is the unique number a such that L(a) = 1.

then you prove that since the area function for 1/x (taken from 1 to x) has derivative 1/x, it lso satisifes those laws. hence it must be a log function!

and by approximating integrals you can show there is some number e between 2 and 3 such that L(e) = 1. and that e is the base.

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mathwonk

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proof: fix y and look at both sides of the equation as functiosn of x. and take their derivatives.

you get (1/xy)(y) on the left by the chain rule, and 1/x on the right, but these are equal.

so the fucntions L(xy) and L(x)+L(y) differ by a constant. but pulgging in x=1, gives L(y) = L(y), so in fact these functions are dead equal.

QED.

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No they don't and no they don't. Actually the lower limit in the integral is 1, thereforeMany books define a "new function" by

[tex]log(x)= \int_0^x \frac{1}{t} dt[/tex]

and then prove that this is, in fact, the usual natural logarithm function.

[tex] \ln x=\int_{1}^{x} \frac{1}{t}{}dt [/tex].

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HallsofIvy

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Ouch!!

(I shall go hide my head!)

(I shall go hide my head!)

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mathwonk

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picky, picky, it was only off by a constant of integration (of + infinity).

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This is interesting I always just asumed it was because [tex]x^{-1} = \frac {1}{0}x^0 =\frac{1}{0}[/tex] or undefined? which of course is nonsense it obviously = [tex] \frac{1}{x}[/tex] or as said a log function, Nice to see there are other reasons.

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disregardthat

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What does the integral mean, does anyone have a link to a proper explanation of it's function?

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arildno

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Eeh??What does the integral mean, does anyone have a link to a proper explanation of it's function?

Whatever are you talking about?

We have defined, for any non-zero, positive x, the function values to be given by a definite integral:

[tex]\log(x)=\int_{1}^{x}\frac{dt}{t}[/tex]

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It actually means sum, and is a large old fashioned S, essentially a definite integral is simply that:What does the integral mean, does anyone have a link to a proper explanation of it's function?

[tex] \int_a^b (F)b-(F)a.[/tex]

It is in a way simillar to summation but different enough to warrant it's own symbol.

In calculus, the integral of a function is an extension of the concept of a sum. The process of finding integrals is called integration. The process is usually used to find a measure of totality such as area, volume, mass, displacement, etc., when its distribution or rate of change with respect to some other quantity (position, time, etc.) is specified. There are several distinct definitions of integration, with different technical underpinnings. They are, however, compatible; any two different ways of integrating a function will give the same result when they are both defined.

The term "integral" may also refer to antiderivatives. Though they are closely related through the fundamental theorem of calculus, the two notions are conceptually distinct. When one wants to clarify this distinction, an antiderivative is referred to as an indefinite integral (a function), while the integrals discussed in this article are termed definite integrals.

The integral of a real-valued function f of one real variable x on the interval [a, b] is equal to the signed area bounded by the lines x = a, x = b, the x-axis, and the curve defined by the graph of f. This is formalized by the simplest definition of the integral, the Riemann definition, which provides a method for calculating this area using the concept of limit by dividing the area into successively thinner rectangular strips and taking the sum of their areas

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mathwonk

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this makes the region under its graph a finite sequence of rectangles.

the integral of a (positive) step function is the sum of the areas of those rectangles.

given any positive function f, consider all positive step functions smaller than f.

the lower integral of f is the smallest number not smaller than any of the integrals of those smaller step functions.

the upper integral is similar, and if they are equal, that common number is the integral.

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can u show me a complete formal proof mathwonk?cuz i find wat u written a bit vague?

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mathwonk

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everyone says this.

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mathwonk

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if you can find an infinitem number of smaler functons, all of whose integrals you know, and getting really close to your fucnton, you may be able tod efine your integrl as the limit of those integrals.

there is nothing here needing proof, this is a definition. unless you have some other definition, then one could prove they give the same number.

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any page that illustrates what ur sayin formally?

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