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In connection with this question about computing integrals of the form $$\mathcal{I}_n=\int_0^\infty f_n(x)\,\mathrm{d}x=\int_0^\infty \frac{\mathrm{d}x}{e^x+x^n}$$ I noticed an interesting trend while plotting the integrand for $n\to\pm\infty$.

On the left, the blue curve is a plot of the integrand with $n=-1000$; on the right, $n=1000$. The orange dashed curve is $e^{-x}$.

enter image description here

I would think $$\lim_{n\to-\infty}\mathcal{I}_n=\int_0^\infty e^{-x}\,\mathrm{d}x=1\\ \lim_{n\to\infty}\mathcal{I}_n=\int_0^\infty 0\,\mathrm{d}x=0$$ yet the plots suggest otherwise; namely, that $$\lim_{n\to-\infty}f_n(x)=e^{-x}\theta(x-1)\\ \lim_{n\to\infty}f_n(x)=e^{-x}\theta(1-x)$$ where $\theta(x)$ is the Heaviside step function, which in turns suggests $$\lim_{n\to-\infty}\mathcal{I}_n=\frac{1}{e}\\ \lim_{n\to\infty}\mathcal{I}_n=1-\frac{1}{e}$$ If I had to guess, my "intuition" of interchanging the limit and integration fails, but what's the precise reason as to why?

Community
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user170231
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2 Answers2

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The part of the integration range where $x<1$ is what is getting you. If $x$ is small, $x^n$ as $n\rightarrow -\infty$ cannot be ignored, as you have done in your first limit. It actually becomes very large compared to the exponential.

bob.sacamento
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Pick any $d\in (0,1)$ and any $D>1+d.$ The sequence $(f_n)_n$ converges uniformly to $e^{-x}$ on $[0,1-d]$ and uniformly to $0$ on $[1+d,D],$ while $x\in [1-d,1+d]\implies 0<f_n(x)<e^{1-d}$ for all $n\in N.$ Now for any $r>0$ take $n$ large enough that $$m>n\implies (\forall x\in [0,1-d]\;(|f_m(x)-e^{-x}|<r)\;\land (\forall x\in [1+d,D]\;(f_m(x)<r)).$$ Then for $m>n$ we have $$\int_0^{1-d}f_m(x)dx= (1-d)r s_1+\int_0^{1-d}e^{-x}dx$$ where $s_1\in [0,1],$ and we have $$\int_{1-d}^{1+d}f_m(x)dx= 2 d s_2 e^{1-d}$$ where $s_2\in [0,1],$ and we have $$\int_{1+d}^D f_m(x)dx= (D-(1+d))rs_3$$ where $s_3\in [0,1],$ and we have $$0<\int_D^{\infty} f_m(x)dx<\int_D^{\infty}D^{-m}dx=D^{1-m}/(m-1).$$ For brevity let $F_n=\int_0^{\infty}f_n(x)dx$ and $J=\int_0^1e^{-x}dx.$ Keeping $d,D,$ and $r$ fixed and letting $n\to \infty,$ we have $$\lim_{n\to \infty}\sup_{m>n}|F_m-J|\leq A+B+C$$ where $A=(1-d)r$ and $B=2 d e^{1-d}+\int_{(1-d)}^1e^{-x}dx$ and $C=(D-(1+d))r.$ Now $r$ is chosen independently of $d$ and $D$ ( That is, we chose $d,D$ and then what followed held for every $r>0$), so we can keep $d,D$ fixed and let $r\to 0.$ This gives $$\lim_{n\to \infty}\sup_{m>n}|F_m-J|\leq B.$$Finally,as $d$ can be any member of $[0,1]$ we can let $d\to 0$ which causes $B\to 0$.

DanielWainfleet
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