Consider the following integral:

\begin{align} F(x) = \lim_{n\rightarrow\infty}\int f_n(x)dx = \lim_{n\rightarrow\infty}\int\frac{x^n+1}{x^n+2}dx \end{align}

How does one evaluate this integral explicitly? I have considered the uniform convergence criterion for the exchange of limit and integration. If we do this, the convergence of $f_n$ depends on the domain in an interesting way:

\begin{align} \lim_{n\rightarrow\infty} f_n(x) = \left\{\begin{array}{l}1\qquad &x\lt-1\\ \text{Does not exist }\qquad &x=-1\\ 1/2\qquad &x\in(-1,1)\\ 2/3\qquad &x=1\\ 1\qquad &x\gt 1\end{array}\right\} \end{align}

My question is then, How can we interpret the indefinite integral? Do we simply define a piecwise continuous function for this while ignoring the two points $\pm$1? or am I way off base here?? A naive capitulation to Wolfram suggests a Hypergeometric function is involved?? I would like to define the following:

\begin{align} F(x) = \left\{\begin{array}xx\qquad &x\in(-\infty,-1)\bigcup(1,\infty)\\ x/2\qquad &x\in (-1,1)\end{array}\right\} \end{align}

What is wrong with this idea? Thx

  • $\begingroup$ Integrate first, then take the limit. The fact that you converge to different limit points for different $x$ values turns out not to matter. $\endgroup$ – Riemann1337 Apr 10 '14 at 4:04
  • $\begingroup$ It would be better you rewrite your question with a definite integral. $\endgroup$ – Felix Marin Apr 10 '14 at 4:33

Another approach with the Heaviside function :

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Note : Concerning the integral (not at limit), i.e.: $F(X,n)$, in case of $X<0$ and $n$ odd it is understood as the Cauchy principal value.

  • $\begingroup$ Nice way to look at it. Thanks $\endgroup$ – JEM Apr 10 '14 at 13:12

A different manner to answer to the question is to show a graphical representation to the function $f(x)$ for various values of $n$ and a graphical representation of the respective integrals $F(X)$. ( The old saw that “One little picture says more than a long speech”)

This makes more understandable the behavior of the function and integral for $n$ tending to infinity. Of course the aim of this approach is not to give a formal and due proof. Nevertheless, well understanding the behavior is a valuable help for further attemp to built a formal proof.

enter image description here

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  • $\begingroup$ Cauchy principle value sounds like the result is multi valued somehow. Is this the case and if so how is the "principle" value singled out? $\endgroup$ – JEM Apr 18 '14 at 20:49
  • $\begingroup$ The Cauchy principal value allows an extension of the theory of integration in some particular cases of improper integrals: mathworld.wolfram.com/CauchyPrincipalValue.html It must not be confused with the "Principal Value" of multivalued functions in the field of complex analysis. The confusing terms of "Cauchy Principal Value" and of "Principal Value" are not related. $\endgroup$ – JJacquelin Apr 18 '14 at 21:46

The indefinite integral is given by

$$ \int\frac{x^n+1}{x^n+2}dx = x-\frac{x}{2}{}_2F_1\left(1,\frac{1}{n}, 1+\frac{1}{n}, -\frac{x^n}{2} \right). $$ In the limit, we have $$ \lim_{n\rightarrow\infty}{}_2F_1\left(1,\frac{1}{n}, 1+\frac{1}{n}, -\frac{x^n}{2} \right)=1, $$ independent of the value of $x$ (the fourth argument can limit to $0$, $\pm 1$, or $\pm\infty$ and the limit of the hypergeometric function is the same in all cases), so the solution is $$ \lim_{n\rightarrow\infty}\int\frac{x^n+1}{x^n+2}dx = \frac{x}{2}. $$

  • $\begingroup$ Yes, I saw this with Mathematica. Took the limit as $n\rightarrow\infty$ and observed the hypergeometric function go to 1 no matter what my $x$ value was. However, I guess I'm looking for a somewhat more Euclidean geometric explanation if possible. Thx $\endgroup$ – JEM Apr 10 '14 at 4:39
  • $\begingroup$ Ah, don't bump me down then if you agree with the solution. Just wait for another solution that you like and select that as your preferred answer. $\endgroup$ – Riemann1337 Apr 10 '14 at 4:40
  • $\begingroup$ I didn't bump you Riemann. Btw, integrating first is an elementary result? $\endgroup$ – JEM Apr 10 '14 at 4:42

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