Where does $\int_0^1 \frac{e^{xy}-1}{y}dy$ converge The question is for what $x$ does it converge?
My try:
$$\int_0^1 \frac{e^{xy}-1}{y}dy = \int_0^1e^{xy}y^{-1} - \int_0^1 y^{-1} dy = \infty +...$$
Does that mean it doesn't converge for any $x$, since $\lim_{y\rightarrow 0}\ln(y) = -\infty$.
In solutions it says otherwise.
EDIT:
$$\lim_{y\rightarrow 0} \frac{e^{xy}-1}{y} = \infty$$
$$\lim_{y\rightarrow 1} \frac{e^{xy}-1}{y}  = e^x-1$$
 A: Termwise integration of the integrand expanded as a Taylor series in $y$ gives
$$\int_0^1\frac{e^{xy}-1}y\,dy=\int_0^1\sum_{n=1}^\infty\frac{x^ny^{n-1}}{n!}\,dy=\sum_{n=1}^\infty\int_0^1\frac{x^ny^{n-1}}{n!}\,dy=\sum_{n=1}^\infty\frac{x^n}{n\cdot n!}$$
which clearly converges everywhere by (e.g.) the ratio test.
A: Note that
$$\lim_{t\to 0}\frac{e^t-1}{t} = e^0 = 1$$
For $x=0$ nothing is to show. For fixed $x\neq 0$ you get
$$\int_0^1\frac{e^{xy}-1}{y}dy = x \int_0^1\frac{e^{xy}-1}{xy}dy$$
where $\frac{e^{xy}-1}{xy}$ is continuous for $y \in (0,1]$ and can be continuously extended into $y=0$. Hence the integral converges for all $x \in \mathbb{R}$.
A: Here is a "general" procedure for testing convergence of the integral of some function on an interval.
So let's say you have, $\int_a^b f_x(y)dy$. That is, $y$ is varying between $a$ and $b$, and $x$ is an external parameter which affects the function $f$. What you want to do, is analyse for what values of $x$ does $\int_a^b f_x(y)dy$ exist. Potentially, $a,b$ can be (positive/negative) infinity as well.

Usually, two common theorems are used for the purpose of testing convergence, at least in the beginning. I'll state them.

*

*$\textrm{Convergence for bounded functions}$ : On a closed and bounded interval $[a,b]$, any bounded function $f_x$ has a bounded integral $\int_a^b f_x(y)dy$. Furthermore , if $f_x$ is continuous on $[a,b]$ then it is bounded on $[a,b]$ and hence has a bounded integral on this interval.


*$\textrm{Equivalence/Comparison Principle}$ : Let $f,g$ be two functions on an interval (possibly unbounded, can be open or closed on either side) $[a,b]$ with the following property : there exist constants $C_1,C_2>0$ such that $C_1 f(x) \leq g(x) \leq C_2f(x)$ for $x \in [a,b]$. Then, $\int_a^b f(x)dx$ exists, if and only if $\int_a^b g(x)dx$ exists.


*$\textrm{Break-up Principle}$ : Suppose $[a,b]$ is broken into various subintervals $[a_1,b_1],...,[a_n,b_n]$ such that $\int_{a_i}^{b_i} f_x(y)dy$ is finite on each of these subintervals. Then, $\int_a^b f_x(y)dy$ exists. On the other hand, if even one of the integrals is infinite or doesn't exist, then $\int_a^b f_x(y)dy$ does not converge.

Therefore, suppose you have an integral like $\int_a^b f_x(y)dy$ whose convergence you'd like to test, you should ideally see if one can use the principles mentioned above to test convergence.
I outline the following strategy for this purpose. This strategy works for most elementary questions, but NOT in general. If I had to be more precise : this doesn't work when the integral is "oscillatory" in nature and the integral is kind of moving up and down without converging. In those cases, more care is required.

*

*Let the interval be $[a,b]$ and suppose you're studying the convergence of $\int_a^b g(x)dx$.


*Find the points in $(a,b)$ where $g$ is "infinite" i.e. points $d$ that $\lim_{x \to d} g(x)$ is $\pm \infty$. Put these points together to create a set $S$ of "problem points". To $S$, add the set of endpoints $[a,b]$ (add $d = \pm \infty$ if they are any of the endpoints).


*At each point $d$ in $S$, find suitable functions $h_d(x)$ defined in a locality of $d$ (so for $d=\infty$, find a function defined on $[j,\infty)$ for $j$ large enough) so that $\lim_{x \to d} \frac{g(x)}{h_d(x)}$ is a finite non-zero number.


*Check if, for each $d$ in $S$, the integral of $h_d(x)$ exists over the neighbourhood on which $h_d$ is defined. If it does for every $d$, then the integral $\int_a^b g(x)dx$ is convergent. If even one case fails, then $\int_a^b g(x)dx$ is not convergent.


*An additional, fine points in the strategy : if , at a problem point $d$, $\lim_{x \to d} g(x)$ exists, then $h_d = 1$ can be taken.

We will use this to solve :
$$
\int_0^1 \frac{e^{xy}-1}{y}dy \quad ; \quad f_x(y) = \frac{e^{xy}-1}{y}
$$
We have to check the convergence of this integral for various fixed values of $x$. Remove the trivial case $x=0$ first : in this case, the function is just the zero function. That is, $x \neq 0$ from now on.
We have a huge advantage in this case (and in a lot of cases): on $(0,1)$, the function $f_x(y)$ is continuous, because it's a quotient of continuous functions on this interval. Therefore the "problem" points in this case are the endpoints, $0$ and $1$. From now on, I attempt to follow the strategy : but not rigorously, only roughly. We will be rigorous later on.
Take the problem point $d=1$. We observe that $$
\lim_{t \to 1} f_x(t) = e^x-1
$$
therefore, we take the function $h_1 =1$ in a neighbourhood of $1$. Now, at the problem point $0$,
$$
\lim_{t \to 1} f_x(t) = \lim_{t \to 1} \frac{e^{xt}-1}{t} = x
$$
Therefore, as $x$ is fixed and finite, we take $h_0 = 1$ here as well.
The appropriate neighbourhoods of $0$ and $1$ will be , well, $(-\delta,\delta)$ and $(1-\delta,1+\delta)$ for small enough $\delta$. Are $h_0,h_1$ integrable over their respective intervals? I mean, obviously, right?
Therefore, following the strategy, we expect convergence for all values of $x$. Now, let's be rigorous.

Rigorous proof
Let us collect the facts from the above exploration that we will need. We will assume that $x$ is positive : I leave the proof for $x$ negative as an exercise. The proof for $x$ negative will involve changing the order of some inequalities below, but nothing more than that. The idea remains the same.

*

*$f_x(y)$ is continuous on $(0,1)$.


*$\lim_{t \to 0} f_x(t) = x$.


*$\lim_{t\to 1} f_x(t) = e^x-1$.
Now, we will make the following rigorous conclusions. Watch how we do this : the same argument will recur many times.


*Since $\lim_{t \to 0} f_x(t) = x$, by the quotient rule we know that $\lim_{t\to 0} \frac{f_x(t)}{x} = 1$, and there exists a $\delta_1>0$ small enough such that for $0\leq x<\delta_1$ , $0.5 \leq \frac{f_x(t)}{x} \leq 1.5$.


*Since $\lim_{t \to 1} f_x(t) = e^x-1$, by the quotient rule we know that $\lim_{t \to 1}{f_x(t)}{e^x-1} = 1$. Therefore, there exists a $\delta_2>0$ small enough such that for $1-\delta_2<x\leq 1$, $0.5 \leq \frac{f_x(t)}{e^x-1} \leq 1.5$.


*Since $f_x(t)$ is continuous on $[\delta_1,1-\delta_2]$, it is bounded. Say the constant bounding it is $M$.
We now perform the following break-up.
$$
\int_0^1 f_x(t)dx = \int_0^{\delta_1} f_x(t)dx + \int_{\delta_1}^{1-\delta_2} f_x(t)dt + \int_{1-\delta_2}^1 f_x(t)dt
$$
Let's deal with each of the parts now : by point 4., we have $$
0.5x\delta_1 =0.5\int_0^{\delta_1} xdt \leq  \int_0^{\delta_1} f_x(t)dt \leq 1.5 \int_0^{\delta_1} xdt = 1.5x\delta_1
$$
By point 6., we have :$$
-(1-\delta_2-\delta_1)M = \int_{\delta_1}^{1-\delta_2} (-M)dt \leq \int_{\delta_1}^{1-\delta_2} f_x(t)dt \leq \int_{\delta_1}^{1-\delta_2} Mdt = (1-\delta_2-\delta_1)M
$$
By point 5., we have :
$$
0.5(e^x-1)\delta_2 =0.5\int_{1-\delta_2}^{1} (e^x-1)dt \leq  \int_{1-\delta_2}^{1} f_x(t)dt \leq 1.5 \int_{1-\delta_2}^{1} (e^x-1)dt = 1.5(e^x-1)\delta_2
$$
Combining the three identities above and using the break-up, we obtain : $$
0.5x\delta_1 +(-(1-\delta_2-\delta_1)M)+0.5(e^x-1)\delta_2 \\ \leq \int_0^1 f_x(t)dt \\ \leq 1.5x\delta_0 +((1-\delta_2-\delta_1)M)+1.5(e^x-1)
\delta_2
$$
which proves that the integral exists for all values of $x$.
