Convergence/Divergence $\int_{0}^{1}\frac{1-e^{-x^2}}{x}\,dx$ 
$$\int_{0}^{1}\frac{1-e^{-x^2}}{x}\,dx$$

I have tried to use the comparison test, and tried to used Dirichlet and Abel but the integrand does not fulfil the requirements.  
 A: If you put $x^{2}=u
 $ we get $$\int_{0}^{1}\frac{1-e^{-x^{2}}}{x}dx=\frac{1}{2}\int_{0}^{1}\frac{1-e^{-v}}{v}dv
 $$ now note that $$\frac{1}{2}\int\frac{1-e^{-v}}{v}dv=\frac{1}{2}\left(\log\left(x\right)-\textrm{Ei}\left(-x\right)\right)+C
 $$ where $\textrm{Ei}\left(-x\right)
 $ is the exponential integral, and since $$\textrm{Ei}\left(x\right)=\frac{1}{2}\left(\log\left(x\right)-\log\left(\frac{1}{x}\right)\right)+\gamma+\sum_{k\geq1}\frac{x^{k}}{kk!}
 $$ it is sufficient to note that $$\lim_{x\rightarrow0^{+}}\left(\log\left(x\right)-\textrm{Ei}\left(-x\right)\right)=-\gamma
 $$ hence $$\int_{0}^{1}\frac{1-e^{-x^{2}}}{x}dx=\color{red}{\frac{1}{2}\left(\gamma-\textrm{Ei}\left(-1\right)\right)}.$$
A: The function is positive and bounded on $(0,1]$. (Use L'Hospital on the integrand as $x\to0^{+}$.) Say it's bounded by $M$.
So $$0\leq\int_t^1f(x)\,dx\leq M(1-t)$$
Taking the limit as $t\to0^{+}$, 
 $$0\leq\int_0^1f(x)\,dx\leq M$$
And since the integrand is monotone increasing on $(0,1]$, this is enough to conclude the integral exists.
A: Probaly not very elegant, but, using Taylor expansion $$e^{-x^2}=\sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{2n}$$ the integrand is $$\frac{1-e^{-x^2}}{x}=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n!}x^{2n-1}$$ and you face the problem of an alternating series since $$\int_0^a\frac{1-e^{-x^2}}{x}dx=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n\,n!}a^{2n}$$
A: Put $g(x)=\exp(-x^2)$. Then as $g$ has for derivative at $x=0$ $g^{\prime}(0)=0$, we get that your function $f(x)=\frac{1-\exp(-x^2)}{x}\to 0$ if $x\to 0$. Putting $f(0)=0$, your function is continuous on $I=[0,1]$, hence Riemann-integrable on $I$.
