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$$\int_{0}^{1} \frac{1}{e^{\sqrt{x}}-1}dx$$

Prove it is convergent or divergent.

The main problem I face is how to deal with the expotential function in such a position.

zhen
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    Hint: Taylor development. – TZakrevskiy May 30 '14 at 15:34
  • Sorry, I'm newbie to calculus, can you be more specific? – zhen May 30 '14 at 15:38
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    Show that $e^{\sqrt{x}}\ge 1+\sqrt{x}$. More simply, show $e^t\ge 1+t$ for $t\ge 0$. – André Nicolas May 30 '14 at 15:39
  • If your function is continuous on $[0,1]$, then it's integrable. The only place where the continuity is not obvious, is $x=0$. But we can prove that $\lim_{x\to 0}\frac{1}{e^{\sqrt x}-1}$ by using Taylor development. – TZakrevskiy May 30 '14 at 15:43
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    On a related note, $$\int_0^\infty\frac1{\exp\big(\sqrt[n]x\big)}~dx=n!$$ $$\int_0^\infty\frac1{\exp\big(\sqrt[n]x\big)-1}~dx=n!\cdot\zeta(n)$$ $$\int_0^\infty\frac1{\exp\big(\sqrt[n]x\big)+1}~dx=n!\cdot\eta(n)$$ – Lucian May 30 '14 at 15:52

1 Answers1

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$1+x \leq e^x$ is a well know inequality which can be derived in many ways.

$$\sqrt{x} \leq e^{\sqrt{x}}-1$$

So $$\int_0^1 \frac{1}{e^\sqrt{x}-1}dx \leq \int_0^1 \frac{1}{\sqrt{x}}dx=2 $$ Thus the integral converges.

Rene Schipperus
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