Can we evaluate this limit? $\lim_{a \rightarrow 0^+}a \int_{b}^{-\log a}\frac{e^y}y \, dy$

Question

Does the following limit exist?

$$\lim_{a \rightarrow 0^+}a \int_{b}^{-\log a}\frac{e^y}y \, dy$$

I suspect it might be $0$, but am unable to get anything concrete.

Answer 1

Hint: $$ \lim_{a \rightarrow 0+} a \displaystyle \int_{b}^{-log \ a}\frac{e^y}{y}dy =\lim_{a \to 0^+} \frac{ \int_{b}^{-log \ a}\frac{e^y}{y}dy }{\frac{1}{a}}=\lim_{z \to \infty} \frac{ \int_{b}^{\log z}\frac{e^y}{y}dy }{z}$$

Show that this is of $\infty/\infty$ type and apply L'Hopital.

Answer 2

Too long for a comment.

Sooner or later, you will learn that $$\int \frac {e^y} y \,dy=\text{Ei}(y)$$ which the definition of the exponential integral function.

So $$I=\int_b^{-\log(a)} \frac {e^y} y \,dy=\text{Ei}(-\log (a))-\text{Ei}(b)$$

On the other hand (have a look here), for large values of $x$ $$\text{Ei}(x)\sim \frac {e^x} x \left(1+\frac{1!}x+\frac{2!}{x^2}+\frac{3!}{x^3}+\cdots\right)$$ So, limited to the first term $$\text{Ei}(-\log (a))\sim -\frac{1}{a \log (a)}$$ and $$a I\sim -\frac{1}{ \log (a)}- a\, \text{Ei}(b)$$