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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. –  André Nicolas Apr 24 '12 at 2:15
    
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 $\alpha$... –  J. M. Apr 24 '12 at 3:17
    
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... –  J. M. Apr 24 '12 at 3:23
    
It only seems strange because you aren't computing $C$ (which depends on $r$). Compute $C$ and see what happens. –  Qiaochu Yuan Apr 24 '12 at 5:24
    
See also the second explanation in my answer at math.stackexchange.com/questions/129849/… for another example of this phenomenon. –  Qiaochu Yuan Apr 24 '12 at 5:27
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marked as duplicate by Qiaochu Yuan Apr 24 '12 at 5:25

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1 Answer

Here's a rather broad class of examples. Suppose $F(x)$ is a differentiable function with $F'(x) = f(x)$. Then $\int f(\lambda x)\ dx = \lambda^{-1} F(\lambda x) + C$ for $\lambda \ne 0$. Of course that doesn't make sense for $\lambda = 0$, where $\int f(0x)\ dx = f(0) x + C$ instead.

EDIT: At first sight your example of $\int t^r \ dt$ doesn't appear to fit this paradigm, but if you first do the change of variables $t = e^x$ it becomes $\int e^{(r+1) x}\ dx$ which does.

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