Calculate limit with integral

Question

Hi I have a problem with following limit: $$\lim_{x\rightarrow\infty}e^{-x}\int_{0}^{x}\int_{0}^{x}\frac{e^u-e^v} {u-v}\ \mathrm du\ \mathrm dv$$ as a hint i got that i should use de l'Hospital. So: $$\lim_{x\rightarrow\infty}\frac{\int_{0}^{x}\int_{0}^{x}\frac{e^u-e^v} {u-v}\ \mathrm du\ \mathrm dv}{e^{x}}$$ And now we want to calculate derivative of up and down. But now i am not sure how to calculate derivative from $$\frac{\mathrm d}{\mathrm dx}\left( \int_{0}^{x}\int_{0}^{x}\frac{e^u-e^v} {u-v}\ \mathrm du\ \mathrm dv\right)$$ I will be very glad for help

Answer 1

With the change of variables $\xi=u$ and $\eta=u-v$, that is $u=\xi$ and $v=\xi-\eta$ and $\left|\frac{\partial (u,v)}{\partial (\xi,\eta)}\right|=1$, the integral becomes $$ f(x)=\int_0^x\int_0^x\mathrm e^{\xi}\,\frac{\mathrm e^{\eta}-1}{\eta}\,\mathrm d\xi\,\mathrm d\eta=\underbrace{\int_0^x\mathrm e^{\xi}\,\mathrm d\xi}_{\phi} \cdot\underbrace{\int_{-x}^x\frac{\mathrm e^{\eta}-1}{\eta}\,\mathrm d\eta}_{\psi}=\phi(x)\psi(x) $$ and observing that $\phi(x)=\mathrm e^{x}$ and $\psi(x)=2\sinh x$ we have $$f(x)=\mathrm e^{-x}\phi(x)\psi(x)=2\sinh x\to\infty$$ for $x\to\infty$.

Answer 2

Define $F(x, y)$ by

$$ F(x, y) = \int_{0}^{x}\int_{0}^{y} \frac{\mathrm{e}^u - \mathrm{e}^v}{u - v} \, \mathrm{d}u\mathrm{d}v. $$

Then we want the derivative of $F(x, x)$. To clarify the structure of this function, let us introduce a new function $g$ defined as $g(x) = (g_1(x), g_2(x)) = (x, x)$. Then $F(x, x) = F \circ g$ and we can apply the chain rule:

$$ \frac{\mathrm{d}}{\mathrm{d}x} F(x, x) = \bigg( \left. \frac{\partial F}{\partial x} \right|_{g(x)} \bigg) \frac{\mathrm{d}g_1(x)}{\mathrm{d}x} + \bigg( \left. \frac{\partial F}{\partial y} \right|_{g(x)} \bigg) \frac{\mathrm{d}g_2(x)}{\mathrm{d}x}. $$

By noticing that

$$ \frac{\partial F}{\partial x} = \int_{0}^{y} \frac{\mathrm{e}^x - \mathrm{e}^v}{x - v} \, \mathrm{d}v, \qquad \frac{\partial F}{\partial y} = \int_{0}^{x} \frac{\mathrm{e}^y - \mathrm{e}^u}{y - u} \, \mathrm{d}u, $$

it is straightforward to check that

$$ \frac{\mathrm{d}}{\mathrm{d}x} F(x,x) = 2 \int_{0}^{x} \frac{\mathrm{e}^x - \mathrm{e}^u}{x - u} \, \mathrm{d}u. $$

A further simplification can be made by applying the substitution $t = x - u$, and then

$$ \frac{\mathrm{d}}{\mathrm{d}x} F(x,x) = 2\mathrm{e}^x \int_{0}^{x} \frac{1 - \mathrm{e}^{-t}}{t} \, \mathrm{d}t. $$

Therefore

$$ \lim_{x\to\infty} \frac{F(x,x)}{\mathrm{e}^x} = 2 \int_{0}^{\infty} \frac{1 - \mathrm{e}^{-t}}{t} \, \mathrm{d}t = \infty. $$