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Possible Duplicate:
Why can't you integrate all power functions without a log function?

We know that $f(r,x)=\int_{1}^xt^r dt=\frac x{1+r}+C$ if $r\neq -1$ and $=\log x+C$ if $r=-1$. I find this quite strange. It's a weird singularity right there, where integrating a monomial $x^r$ would "escape" to another class of equations only at $r=-1$.

  1. Is there any explanation for this phenomenon?

  2. What about similar singular behaviors in other such functions?

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    We have $e^{\lambda x}$ at $\lambda=0$. And many relatives with a division by $0$ issue.2012-04-24
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    Consider $x^\alpha=\sum\limits_{k=0}^\infty \binom{\alpha}{k}(x-1)^k$. Integrating that yields $\int_1^x t^\alpha \mathrm dt=\sum\limits_{k=0}^\infty \binom{\alpha}{k}\frac{(x-1)^{k+1}}{k+1}$ From the series point of view, things aren't *too* strange, since $\binom{\alpha}{k}$ is nothing more than a degree-$k$ polynomial in $\alpha$...2012-04-24
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    Consider also the Chebyshev polynomials of the first kind $T_k(x)=\cos(k\arccos\,x)$. There is the integration formula $\int T_k(x)\mathrm dx=\frac12\left(\frac{T_{k+1}(x)}{k+1}-\frac{T_{k-1}(x)}{k-1}\right)$, which very obviously breaks down when $k=1$; however, $T_1(x)=x$, and that function is perfectly integrable...2012-04-24
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    It only seems strange because you aren't computing $C$ (which depends on $r$). Compute $C$ and see what happens.2012-04-24
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    See also the second explanation in my answer at http://math.stackexchange.com/questions/129849/linear-homogeneous-recurrence-relations-with-repeated-roots-motivation-behind-l/129855#129855 for another example of this phenomenon.2012-04-24

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