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Messing around with the idea of point convergence, I found out that the following (principal value of the) integral can be evaluated as follows:

$$\int_b^ax^ndx=\int_b^{+\infty}x^ndx+\int_{-\infty}^ax^ndx$$

You may observe its truth, graphically, for odd $n$ integer, and any $a$, or $b$, because of the symmetry of $x^n$, the $+\infty$ sort of cancels with the $-\infty$.

However, how do I prove that the integrals are indeed true?

It is easy to solve for $n<0$, but as for $n\ge0$, it gets more challenging.

I tried the following:

$$\int_b^{+\infty}x^ndx+\int_{-\infty}^ax^ndx=\lim_{y\to\infty}\frac1{n+1}y^{n+1}-\frac1{n+1}b^{n+1}+\frac1{n+1}a^{n+1}-\lim_{z\to-\infty}\frac1{n+1}z^{n+1}$$

But how to proceed?

Thomas Andrews
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Simply Beautiful Art
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  • For $n \ge 0$ (rather, $n \ge -1$) these integrals diverge and the expressions are no longer meaningful. You can talk about principal value, but that's something else. –  Jan 28 '16 at 01:48
  • @T.Bongers Oh, that's what I meant, the principal values are the same. BTW, what does it mean to be meaningful? – Simply Beautiful Art Jan 28 '16 at 01:50
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    On the first line, you have an expression of the form $\infty - \infty$, which doesn't mean anything. The same issue happens after the evaluation of the integrals too, for some values of $n$. –  Jan 28 '16 at 01:51
  • @T.Bongers Yes, I noticed, but graphically it appears as though the particular $\infty-\infty$ can be "reduced" to $F(a)-F(b)$. – Simply Beautiful Art Jan 28 '16 at 01:53
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    No. The RHS is a difference of two limits, neither of which exists (as real numbers), so the difference is indeed meaningless. The RHS is NOT a single limit; the two limits are necessarily independent, so I would say you can't even try to save this using P-values here. – MPW Jan 28 '16 at 02:37
  • @MPW Then why does it appear graphically as though it does have a total combined P-value of $F(a)-F(b)$? I think that if it appears to exist graphically, it can have a P-value, one that is $F(a)-F(b)\dots$, and even if something doesn't appear to exist graphically (it still diverges), it can still have a P-value. Right? – Simply Beautiful Art Jan 28 '16 at 21:33
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    @SimpleArt: I guess what I mean is that $\lim_{P\to\infty}f(P)+\lim_{Q\to -\infty}g(Q)$ and $\lim_{R\to\infty}[f(R)+g(-R)]$ needn't be the same. If the first exists, they are the same, but not otherwise. The expression with two integrals is the first, the single integral is what you get if you use P-values. They aren't the same. – MPW Jan 29 '16 at 03:06
  • @MPW Ok. I get what you mean. – Simply Beautiful Art Jan 29 '16 at 13:17

1 Answers1

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What this comes down to is whether or not this is an equality:

$$\lim\limits_{y\to\infty} y^{n+1} - \lim\limits_{z\to-\infty} z^{n+1} \mathop{\quad\boxed ?\quad}^\neq_= \lim\limits_{y\to\infty} (y^{n+1}-(-y)^{n+1})$$

The left hand side is the difference of two non-finite limits, while the right hand side is seemingly equal to zero for odd $n$.

However, can we make the substitution of $-y$ for $z$ followed by subtracting indefinite limits?

[edit:] Hint: What would happen if we tried the substitution $-2y$ for $z$? [/edit]

Graham Kemp
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