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I have to prove the continuity and find the limit $(x,y)\to(1,1)$ of $f:D\to R$, where $$D=\{(x,y)\in R^2:|x|\neq|y|\}$$ and $$ f(x,y) = \frac{1}{y^2-x^2}\int_x^y\ln(e+e^t)dt.$$

I don't know whether my way of thinking is correct and I am a little stuck on the limit.

Continuity

I have had a few thoughts:

  • The multiplication of continuous functions is continuous, thus if $\int g(x)$ is differentiable and the fraction is continuous, it should be continuous.
  • To be integrable, the integral of an absolute value of the function (the derivative) should have a finite limit.

My steps

$$ f(x,y) = \frac{1}{y+x}\cdot\frac{1}{y-x}\int_x^y\ln(e+e^t)dt$$

Since we can bound the integral by:

$$\int_x^y\ln(e+e^t)dt<\int_x^y\ln(e^t)dt=\int_x^ytdt$$

Which is convergent on a given interval, thus $g(x)$ is integrable and continuous

Additionally, from the intermediate value theorem we have: $$\frac{1}{y-x}\int_x^y\ln(e+e^t)dt=\ln(e+e^c)~~~~\text{for some}~~c\in(x,y)$$

Therefore: $$ f(x,y)=\frac{1}{y+x}\cdot\ln(e+e^c) $$

Since $|x|\neq|y|$ the function does not take zero value in the denominator, thus this multiplication of functions is continuous.

I am still learning and I would really appreciate if you would point out my mistakes or present a more accurate solution.

Limit

I have also tried spheric coordinates, but I can't see nothing helpful in this case:

$$ \frac{1}{y^2-x^2} = \frac{1}{r^2(\sin^2\theta-\cos^2\theta )} $$

One of my thoughts was that I would have to find the limit of the integral and of the fraction or calculate the integral and combine with the fraction, however I decided to test the fraction first.

$$ h(x,y)_{(x,y)\to(1,1)} = \frac{1}{y^2-x^2}$$

I check two sequences: $a_n=(1, \frac{1}{n})$ and $ b_n=(\frac{1}{n},1)$ with $n\to\infty$ and get:

$$ \lim_{n\to -\infty} a_n=\lim_{n\to \infty}\frac{n^2}{n^2-1}= \infty\\ \lim_{n\to \infty} b_n=\lim_{n\to \infty}\frac{n^2}{1-n^2} = -\infty~~~~\text{for big}~~n$$

For this function the limit does not exist since we can show two sequences with different limits.

Again, I am not sure whether my solution presents the correct way of solving. I would like to ask you for some guidance.

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There are some problems. First of all, when you use the intermediate value theorem, you find a $c$ for specific $x$ and $y$. So you can't conclude that $f(x,y)=\frac{1}{y^2-x^2}\ln(e + e^c)$ for all $x,y$.

When calculating the limit, you only conclude that $\frac{1}{y^2 -x^2}$ does not have a limit as $(x,y)$ approaches $(1,1)$, which is correct. But this doesn't prove that $f(x,y)$ doesn't have a limit.

To prove the continuity of $f(x,y)$, it is easier to prove continuity of one variable and keeping the other constant for both $x$ and $y$. This will imply continuity of the function of both variables. So let $y$ be given, and define $g(x)=f(x,y)$. On the domain $D$, it is a product of two continuous functions. $\frac{1}{y^2 -x^2}$ is continuous since it is a composition of continuous function. $\int_x^y \ln(e + e^t)dt$ is differentiable as a function of $x$, since $\ln(e + e^t$) is continuous, so in particular it is continuous. Similar arguments proves that the function $h(y)=f(x,y)$ is continuous, so we conclude that $f(x,y)$ is continuous.

The limit at $(1,1)$ is a bit more tricky, let me get back on that when I get the spare time and some paper.

Edit: Now for the limit. This is a little difficult, and I'm not sure you've seen all the techniques used here. Anyway, first notice that

$\ln(e+e^t)=\ln(e(1+e^{t-1})) = 1 + \ln(1+e^{t-1})$

$\ln(1+x)$ has a particularly nice series expansion, which we can now use. First notice that

$\int_x^y 1 + \ln(1+e^{t-1})dt = \int_{x-1}^{y-1} 1 + \ln(1+e^{t})dt$

So let us look at that last integrand. From the series expansion, we get

$1+\ln(1+e^t)=1+e^t-\frac{e^{2t}}{2}+ \frac{e^{3t}}{3}-...$

Integrating this term by term we get

$\int_{x-1}^{y-1} 1 + \ln(1+e^{t})dt=\left[ t + e^t-\frac{e^{2t}}{4}+...\right]_{x-1}^{y-1}$

$=y-x+e^{y-1}-e^{x-1}-...$

Notice that $\frac{e^{y-1}-e^{x-1}}{y-x}$ and higher powers tends to $0$ when $(x,y)$ goes to $(1,1)$ (you can use L'hopital to show this). Therefore, then only part that doesn't go to zero when divided by $y-x$ is $y-x$. In total we get

$\lim_{(x,y) \rightarrow (1,1)}\frac{1}{y^2-x^2}\int_x^y \ln(e+e^t)dt = \lim_{(x,y) \rightarrow (1,1)}\frac{1}{y+x}\frac{1}{y-x}(y-x)=\frac{1}{2}$

This got a bit advanced fast, and I'm not 100% if all the calculations are correct, but try to see if you can make heads or tails of it.

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  • $\begingroup$ Thank you for such a detailed comment. I would appreciate your way to solve the limit. It is very interesing that the limit can still exist. $\endgroup$
    – Funny
    Oct 21 '20 at 14:32
  • $\begingroup$ Thank you for your answer. I really appreciate, that you cared about my understanding; this proof is clearly explained. However, I am quite confused how you got the division by $y-x$ in the series - quite unexpected for me. Additionally, could you tell me whether I predicted the convergence of an integral correctly? $\endgroup$
    – Funny
    Oct 21 '20 at 19:23
  • $\begingroup$ The reason I am dividing by y-x is because we need to multiply the integral with $\frac{1}{y^2-x^2}$ which you yourself pointed out is equal to $\frac{1}{y+x}\frac{1}{y-x}$. The first part of that product goes to $\frac{1}{2}$ as $(x,y)$ goes to $(1,1)$, so there is no problem, but the other doesn't have a limit. Therefore, I tried to multiply that with the integral (in its series form) to see if I got something convergent. $\endgroup$ Oct 22 '20 at 12:19
  • $\begingroup$ I think you're a little off when you predict the convergence of the integral. Just look at the integrand $\ln(e+e^t)$. This is a continuous function, so by the fundamental theorem of calculus you get that it is integrable on the interval $[x,y]$ (for $y>x$), and that the integral function of $x$ and $y$ is continuous. $\endgroup$ Oct 22 '20 at 12:22

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