Let $f_1(x)=e^x$ and define recursively $$ f_{n+1}(x)=(e^x)^{f_n(x)}. $$

Does the following limit exist? $$ \lim_{n\to \infty}\int \dfrac{f_n(x)}{\pi^n}dx $$

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    $\begingroup$ What are the limits of integration? $\endgroup$ – Brian Tung Jun 7 '16 at 23:16
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    $\begingroup$ What a strange problem. Well you can start with ignoring the integral sign alltogeather. It makes little difference here. $\endgroup$ – Winther Jun 7 '16 at 23:41
  • $\begingroup$ @Winther: How do you figure? I guess I'm not sure how you're interpreting the question. $\endgroup$ – Brian Tung Jun 7 '16 at 23:46
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    $\begingroup$ @BrianTung The integrand does not converge for any $x > 0$. I interpreted it as $e^{xe^{xe^x}}$ $\endgroup$ – Winther Jun 7 '16 at 23:52
  • $\begingroup$ Friends,friends,I didn't add limits,because I created this problem and ,I want to learn ,which method we can use. $\endgroup$ – Anıl B.C.T. Jun 8 '16 at 4:51

Assuming you are talking about limits of functions (the limit of antiderivatives to be exact), note that $(a^b)^c = a^{bc}$. Thus we are finding

$$ f(x) = \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{1}{\pi^n} \int e^{xe^{nx}} dx $$

Fix a constant such that $f_n(x) = 0$ for each $n$, so that $f_n(x) = \frac{1}{\pi^n} \int_0^x e^{xe^{nx}} dx$. For $x > 1$, $e^{xe^{nx}} > e^{e^n}$. Thus on these values $f_n(x) \geq \frac{1}{\pi^n} (x-1) e^{e^n}$, which tends to infinity as $n$ tends to infinity. One can refine this method to show that the series diverges (converges to infinity) on $(0,\infty)$.

For $x < 0$, we note that $e^{xe^{nx}}$ is a decreasing sequence, hence the Lebesgue monotone convergence theorem implies that $\lim_{n \to \infty}\int_0^x \frac{e^{xe^{nx}}}{\pi^n}$ exists, and is equal to $\int_0^x \lim_{n \to \infty} \frac{e^{xe^{nx}}}{\pi^n}$. Now $e^{xe^{nx}}$ converges to zero for $x < 0$, for $nx \to -\infty$, hence $e^{nx} \to 0$, hence $\frac{e^{xe^{nx}}}{\pi^n} \to 0$. Thus for $x < 0$, $f(x) = 0$.

In conclusion, we find $f = \infty \cdot \chi_{(0,\infty)}$. To cheat a little, checking our answer is right, we look at the graph below for a large value of $n$:

enter image description here

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    $\begingroup$ Are you assuming that OP intends the exponentiation to left-associate? I would have assumed right-association. $\endgroup$ – Brian Tung Jun 7 '16 at 23:45
  • $\begingroup$ That would make sense, seeing as how he didn't simplify above $\endgroup$ – Jacob Denson Jun 7 '16 at 23:46

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