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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. :)

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    $\begingroup$ Consider $$g(x) = \frac{e^{\sin x} - 1}{x}.$$ How does $g$ behave near $0$? $\endgroup$ – Daniel Fischer Nov 8 '17 at 16:43
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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 $x=1$.

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\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*}

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