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Let $$F(x)= \int_1^x \frac{1}{2 \sqrt{t}-1}dt $$ for all $x\geq1$. Then if $c>0$, there is a unique solution to the equation $F(x)=c$, $x>1$.

I calculated the integral but it didnt seem to help. What approach should I take?

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    $\begingroup$ I'm worried by integrability at $t=1/4$, since $2\sqrt{\epsilon+1/4}-1 = 2\epsilon + O(\epsilon^2)$. Is it supposed to be a Cauchy PV integral ? $\endgroup$ – Stop hurting Monica May 2 '14 at 7:53
  • $\begingroup$ Sorry it is supposed to be 1 I edited it $\endgroup$ – Paul Malinowski May 2 '14 at 8:00
  • $\begingroup$ Ok, now do you want to prove there is a solution for all $c>0$, or to have a closed form for the solution? To prove there is a solution, just notice the integrand is positive (so the primitive is increasing), and by comparison with the integral of $1/\sqrt{x}$, the primitive is obviously not bounded when $x\rightarrow +\infty$. Then apply intermediate value theorem, since the primitive of a continuous function is continuous. $\endgroup$ – Stop hurting Monica May 2 '14 at 8:06
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Note that $F'(x) = \frac{1}{2\sqrt{x} - 1}$. Since $x > 1$, what does that tell you about $F(x)$? What is $F(0)$? What is $\lim_{x \to \infty} F(x)$? How does the intermediate value theorem help?

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  • $\begingroup$ I don't understand. If $x>1$ then $0<F'(x)<1$, so $F$ is strictly increasing, and thus one-to-one? $\endgroup$ – Paul Malinowski May 2 '14 at 7:21
  • $\begingroup$ And so the inverse exists and is also one-to-one, and the result follows? is that right?' $\endgroup$ – Paul Malinowski May 2 '14 at 7:23
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    $\begingroup$ Strictly increasing is one thing, but you also have to show that $F(x)$ is unbounded, since there are strictly increasing functions which are still bounded above and hence don't reach every positive real value. From there you can conclude that your function is one-to-one and has an inverse, if you feel that's obvious enough. $\endgroup$ – FlagCapper May 2 '14 at 7:32
  • $\begingroup$ And $F(0) = -3/2$? $\endgroup$ – Paul Malinowski May 2 '14 at 7:34
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Since the answers you received perfectly clarify your problem, le me focus on the solution. $$F(x)= \int_0^x \frac{1}{2 \sqrt{t}-1}dt=\frac{1}{2} \left(2 \sqrt{x}+\log \left(2 \sqrt{x}-1\right)-1\right)+\frac{1}{2} (1-i \pi )$$ and the unique solution of $F(x)=c$ is given by $$x=\frac{1}{4} \left(W\left(-e^{2 c-1}\right)+1\right)^2$$ where appears Lambert function.

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  • $\begingroup$ It would be interesting to see the derivation. At least to understand where this $i$ come from, since everything else is real... $\endgroup$ – Stop hurting Monica May 2 '14 at 7:45

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