Show that $\int_0^\infty \frac{e^{\sin x}-1}{\sqrt{x}x}dx$ converges I have to show that the above improper integral is convergent and I'm stuck. But this is my work so far:


*

*Split it: $$\int_0^\infty \frac{e^{\sin x}-1}{\sqrt{x}x}dx = \int_0^\frac{\pi}{2} \frac{e^{\sin x}-1}{\sqrt{x}x}dx + \int_\frac{\pi}{2}^\infty \frac{e^{\sin x}-1}{\sqrt{x}x}dx$$
The split point $\frac{\pi}{2}$ is for obvious reasons. 

*Then I work on the right term: $$\frac{e^{\sin x}-1}{\sqrt{x}x} \leq \frac{e}{\sqrt{x}x}$$

*$$\lim_{B \to \infty}\int_\frac{\pi}{2}^B \frac{e}{\sqrt{x}x}dx = \sqrt{\frac{2}{\pi}}2e$$ So that part of the integral is convergent. 


But the left term does not yield as easily. Since $\sin x \approx x$ when $x \to 0$, I'm doing the following comparison:
$$\frac{e^{\sin x}-1}{\sqrt{x}x} \leq \frac{e^x-1}{\sqrt{x}x}$$
So now I "only" have to show that
$$\int_0^\frac{\pi}{2} \frac{e^x-1}{\sqrt{x}x}dx$$
is convergent. But I have worked for a long while on it and I'm getting nowhere so please help me. :)
 A: That the second part is convergent can be proved more straightforwardly. Use
$$e^{\sin x}-1\leq e-1.$$
Therefore
$$\int_{1}^\infty\frac{e^{\sin x}-1}{x\sqrt{x}}dx\leq\int_{1}^\infty\frac{e-1}{x^{3/2}}dx<\infty$$
is convergent. To show that the first part also converges, use
$$\int_0^{1}\frac{e^x-1}{x\sqrt{x}}dx<\int_0^{1}\frac{Cx}{x\sqrt{x}}dx<\infty,$$
where $C=e-1$ is the slope of the line segment connecting $(x,e^{x}-1)$ at $x=0$ and $x=1$.
A: \begin{align*}
\frac{e^{x}-1}{x^{3/2}}=\frac{1}{x^{1/2}}+\sum_{k=2}\frac{1}{k!}x^{k-3/2}
\end{align*}
we know that 
\begin{align*}
\int_{0}^{\pi/2}\frac{1}{x^{1/2}}dx<\infty
\end{align*}
and 
\begin{align*}
\sum_{k=2}\frac{1}{k!}\int_{0}^{\pi/2}x^{k-3/2}dx=\sum_{k=2}\frac{1}{k!}\frac{1}{k-(1/2)}x^{k-(1/2)}\bigg|_{x=0}^{x=\pi/2}=\left(\frac{2}{\pi}\right)^{1/2}\sum_{k=2}\frac{1}{k!}\frac{1}{k-(1/2)}\left(\frac{\pi}{2}\right)^{k}
\end{align*}
but 
\begin{align*}
\sum_{k=2}\frac{1}{k!}\frac{1}{k-(1/2)}\left(\frac{\pi}{2}\right)^{k}\leq\sum_{k=2}\frac{1}{k!}\left(\frac{\pi}{2}\right)^{k}<\infty.
\end{align*}
