Confusion with calculating a limit $F(x):=\int_{x}^{2x}e^{-2t}t^{-1}dt$

Question

The question is given in the following way:

Define $F:(0,\infty)\rightarrow \mathbf{R}$ by $$F(x):=\int_{x}^{2x}e^{-2t}t^{-1}dt$$ Determine whether or not $\lim_{x\rightarrow 0^+}F(x)$ exists.

In the first step I write the series of $e^{-2t}$ : $$e^{-2t}=1-(2t)+\frac{(2t)^2}{2!}-...$$ Then $$\frac{e^{-2t}}{t}=\frac{1}{t}-2+\frac{2^2t}{2!}- ...$$ Then I integrate the series termwise $$\int_{x}^{2x}e^{-2t}t^{-1}dt=\int_{x}^{2x}\frac{1}{t}dt-\int_{x}^{2x}2dt+...$$ So, if we take the $\lim_{x\rightarrow0^+}$ and integrating termwise I get $ln(2).$

Process-wise, it seems right, but many things are coming into my mind, such as how the infinite terms are valid for termwise integration and how we are interchanging the limit. Please help me to write it freshly. Thank you beforehand!!!

You can just do $\int_x^{2x}\frac{e^{-2t}-1}t,dt+\int_x^{2x} \frac1t,dt$. The first term has an integrand that is continuous on $\Bbb R$, therefore it goes to $0$ as $x\to 0$. The second term goes to (actually, is) $\ln 2$. — Sassatelli Giulio, Dec 10 '24 at 08:10

Kavi Rama Murthy · Accepted Answer · 2024-12-10T08:39:52.547

6

The substitution $s=t/x$ gives $F(x)=\int_1^{2} e^{-2sx}\frac 1 sds$. This tends to $\int_1^{2} \frac 1 sds=\ln 2$. Justification: DCT.

If you are not familar with DCT, just note that $e^{-2sx} \to 1$ uniformly as $x \to 0+$.

edited Dec 10 '24 at 08:39

answered Dec 10 '24 at 08:09

Kavi Rama Murthy

359,332

1

Shouldn’t it be $e^{-2sx}$? – Integreek Dec 10 '24 at 08:25
@MathGuy Of course! Thanks for pointing out. – Kavi Rama Murthy Dec 10 '24 at 08:27
What is DCT in simple terms? On the Internet, I see Dominated Convergence Theorem which is defined in terms of complex terms like Lebesgue integrable functions, measure space which I cannot understand. – Integreek Dec 10 '24 at 08:32
@MathGuy Tags used by OP include Real Analysis an Integration. So, I believe OP knows about DCT. I have also added a proof without DCT. – Kavi Rama Murthy Dec 10 '24 at 08:37
Ok, so as much as I could understand about DCT from here, it basically allows us to write the limit of an integral as the integral of a limit given that the integrand follows certain conditions(which I am not able to grasp). Am I right? – Integreek Dec 10 '24 at 08:46
Yes, that is right. @MathGuy – Kavi Rama Murthy Dec 10 '24 at 08:50

score 2 · Answer 2 · answered Dec 10 '24 at 08:17

The series expansion

$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$

converges uniformly on all compact subsets of $\mathbb R$, and so in particular

$$e^{-2t}t^{-1}=\frac{1}{t}-2\sum_{n=0}^\infty\frac{(-2t)^n}{(n+1)!}$$

converges uniformly on the compact sets $[x,2x]$ for $x>0$. As uniform convergence allows us to interchange limit and integral, this justifies your termwise integration. Furthermore the resulting series you get (i.e. everything except the first term) is a power series with positive radius of convergence, and in particular it defines a continuous function, which justifies getting rid of it by letting $x\downarrow0$.

Confusion with calculating a limit $F(x):=\int_{x}^{2x}e^{-2t}t^{-1}dt$

2 Answers2