What is so special about α=−1\alpha=-1 in the integral of xαx^\alpha?

Of course, it is easy to see, that the integral (or the
antiderivative) of f(x)=1/x is log(|x|) and of course for
α1 the antiderivative of f(x)=xα is
xα+1/(α+1).

I was wondering if there is an intuitive (probably geometric)
explanation why the case α=1 is so different and why the
logarithm appears?

Some answers which I thought of but which are not convincing:

  1. Taking the limit α=1 either from above or below lead to diverging functions.

  2. Some speciality of the case α=1 are that both asymptotes are non-integrable. However, the antidrivative is a local thing, and hence, shouldn’t care about the behavior at infinity.

Answer

Assume you know that for every \beta the derivative of the function x\mapsto x^\beta is the function x\mapsto\beta x^{\beta-1} and that you want to choose \beta such that the derivative is a multiple of the function x\mapsto x^{\alpha}. You are led to solve the equation \beta-1=\alpha, which yields \beta=\alpha+1. If \alpha=-1, this gets you \beta=0, but then the derivative you obtain is the function x\mapsto 0x^{-1}=0, which is not a nonzero multiple of x\mapsto x^{-1}. For every other \alpha, this procedure gets you an antiderivative but for \alpha=-1, you failed. Or rather, you proved that no power function is an antiderivative of x\mapsto x^{-1}. Your next step might be (as mathematicians often do when they want to transform one of their failures into a success) to introduce a new function defined as the antiderivative of x\mapsto x^{-1} which is zero at x=1, and maybe to give it a cute name like logarithm, and then, who knows, to start studying its properties…

Edit (Second version, maybe more geometric.)

Fix s>t>0 and c>1 and consider the area under the curve x\mapsto x^\alpha between the abscissæ x=t and x=s on the one hand and between the abscissæ x=ct and x=cs on the other hand. Replacing x by cx multiplies the function by a factor c^\alpha. The length of the interval of integration is multiplied by c hence the integral itself is multiplied by c^{\alpha+1}.

On the other hand, if an antiderivative F of x\mapsto x^\alpha is a multiple of x\mapsto x^\beta for a given \beta, then F(ct)=c^\beta F(t) and F(cs)=c^\beta F(s) hence F(ct)-F(cs)=c^\beta (F(t)-F(s)). Note that this last relation holds even if one assumes only that F is the sum of a constant and a multiple of x\mapsto x^\beta.

Putting the two parts together yields c^{\alpha+1}=c^\beta. Once again, if \alpha=-1, this would yield \beta=0, hence F would be constant and the area F(t)-F(s) under the curve x\mapsto x^\alpha from s to t\ge s would be zero for every such s and t, which is impossible since the function x\mapsto x^\alpha is not zero. (And for every \alpha\ne1, this scaling argument yields the correct exponent \beta.)

Attribution
Source : Link , Question Author : Dirk , Answer Author : Community

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