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I am asked to solve this question: 'Which one is the higher order infinitesimals when $\mathbf{x \rightarrow 0 \, \int_0^{\sin x} (1+t)^{\frac{1}{t}} dt}$' or $\mathbf{x^2}$.

I know that:I have to solve $\lim_{x \rightarrow 0}\frac{\int_0^{\sin x} (1+t)^{\frac{1}{t}} dt}{x}$ by using L 'Hospital's rule and compare the result with $x^2$ But I cant find the derivative.

I have recited that the derivative of $\int_{φ(x)}^{ψ(x)}f(t)dt$ (if x is not in f(t)) is $f[φ(x)]φ'(x)-f[(ψ)]ψ'(x)$, without actually knowing how it comes.

But for now, this equation seems impossible since I cannot get $(1+0)^{\frac{1}{0}}$, I don't think $\lim (1+0)^{\frac{1}{0}}$ should be right, since the hint said the $\int_0^{\sin x} (1+t)^{\frac{1}{t}} dt$ ~ $ex$

So what should I do? Is the equation I have recited succeed all the time?

scbz
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    Do you see that the integrand tends to $e$ as $t\to 0$? If yes, then by FTC the desired limit is $e$. – Paramanand Singh Feb 09 '21 at 09:39
  • When $t \to 0$, integrand tends to $e$ and $e$ is a constant which will vanish after the derivation. So that we only consider when $t \to \sin x$. Is this right? – scbz Feb 09 '21 at 09:53
  • I dont think I was right then, but now I know that $\mathcal{O}$ can be used doing integral as well! In this case, just to find the limit and by FTC, the result comes! – scbz Feb 09 '21 at 10:04

2 Answers2

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First note that, as $t\to 0$, $$ (1 + t)^{\frac{1}{t}} = \exp \left( {\frac{1}{t}\log (1 + t)} \right) = \exp \left( {1 + \mathcal{O}(t)} \right) = e\exp \left( {\mathcal{O}(t)} \right) = e + \mathcal{O}(t). $$ Hence $$ \int_0^{\sin x} {(1 + t)^{\frac{1}{t}} dt} = e\sin x + \mathcal{O}(\sin ^2 x) = ex + \mathcal{O}(x^2 ) $$ as $x\to 0$.

Gary
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  • Thanks a lot! Now I know that $\mathcal{O}$ can be used doing integral as well! – scbz Feb 09 '21 at 10:02
  • @scbz Yes, look at https://math.stackexchange.com/questions/116668/how-does-one-integrate-landau-symbols for an example of how to justify it properly. – Gary Feb 09 '21 at 10:06
  • Thanks! I have read that. But one more question, to use Landau Symbol, we should avoid derivation. Is this against L 'Hospital's rule? – scbz Feb 09 '21 at 10:17
  • You cannot differentiate big-$\mathcal{O}$ in general. If your application of L'Hôpital's rule does not involve such maneuvers, you are fine. – Gary Feb 09 '21 at 10:25
  • Oh thanks! Got it! – scbz Feb 09 '21 at 11:05
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I expand on my comment for your benefit.

Let's first analyze the integral $\int_0^{\sin x} (1+t)^{1/t}\,dt$. You should note that the integrand is not defined at $t=0$. However, by definition the symbol $\int_a^b f(x) \, dx$ requires $f$ to be defined in whole interval $[a, b] $.

In such a situation when the integrand is not defined at a finite number of points in the interval of integration, the convention is to define the function at those exceptional points in any manner whatsoever. This does not create any problem because the value of an integral or its existence is not affected by behavior of function at a finite number of points.

Next we define the integrand in question at $t=0$ in such a manner that the function becomes continuous there so that we can take advantage of the Fundamental Theorem of Calculus.

Let $f(0)=e$ and $f(t)=(1+t)^{1/t}$ for $t\neq 0$. Then $f$ is defined in some neighborhood of $0$ and moreover is continuous in that neighborhood.

Now $$\lim _{x\to 0}\frac{1}{x}\int_0^{\sin x} (1+t)^{1/t}\,dt=\lim_{x\to 0}\frac{\sin x}{x}\cdot\frac{1}{\sin x} \int_{0}^{\sin x} f(t) \, dt$$ The first fraction tends to $1$ and next we replace $\sin x$ with $u$ and note that $u$ tends to $0$ with $x$. Hence the desired limit is $$\lim_{u\to 0}\frac{1}{u}\int_0^u f(t) \, dt$$ By FTC this equals $f(0)=e$ as $f$ is continuous at $0$.

Paramanand Singh
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