I have to show that the above improper integral is convergent and I'm stuck. But this is my work so far:

  1. 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.
  2. Then I work on the right term: $$\frac{e^{\sin x}-1}{\sqrt{x}x} \leq \frac{e}{\sqrt{x}x}$$
  3. $$\lim_{B \to \infty}\int_\frac{\pi}{2}^B \frac{e}{\sqrt{x}x}dx = \sqrt{\frac{2}{\pi}}2e$$ 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. :)

  • 1
    $\begingroup$ Consider $$g(x) = \frac{e^{\sin x} - 1}{x}.$$ How does $g$ behave near $0$? $\endgroup$ Nov 8, 2017 at 16:43

2 Answers 2


That the second part is convergent can be proved more straightforwardly. Use

$$e^{\sin x}-1\leq e-1.$$


$$\int_{1}^\infty\frac{e^{\sin x}-1}{x\sqrt{x}}dx\leq\int_{1}^\infty\frac{e-1}{x^{3/2}}dx<\infty$$

is convergent. To show that the first part also converges, use


where $C=e-1$ is the slope of the line segment connecting $(x,e^{x}-1)$ at $x=0$ and $x=1$.


\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*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.