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Is there a solution to definite integral $$\int_{0}^{y} \frac{A + t}{B-t} dt = N$$

Where $A$, $B$, $N$ are constants.

pavybez
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    Please correct your formula, integrals such as $$\int_0^yf(y)dy$$ using the same symbol for the variable of integration and for a bound of the integral, are meaningless. – Did Sep 29 '17 at 06:53
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    @cdummie I cannot, since I do not know what $$\int_0^yf(y)dy$$ can even mean. – Did Sep 29 '17 at 07:14
  • @Did Well, if you are referring to upper bound of integration, as you stated in your previous comment, that should not be the problem, you can simply change it to capital y or something. – cdummie Sep 29 '17 at 07:28
  • @cdummie Then change it to anything you want except $y$ instead of basically cheating on the OP by reproducing their serious mistake as if it was ok. – Did Sep 29 '17 at 07:33
  • @Did I deleted it, since i was unable to edit comment, however it wasn't on purpose, i had no intention to confuse OP. – cdummie Sep 29 '17 at 07:41

1 Answers1

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$$F(y)=\int_0^y \frac{A+t}{B-t}\,dt=\int_0^y \frac{A+B-(B-t)}{B-t}\,dt=(A+B)\int_0^y\frac {dt}{B-t}-\int_0^y dt$$ $$F(y)=-(A+B)\log(1-\frac y B)-y$$ provided that $y<B$.

Now, solving for $y$ $$-(A+B)\log(1-\frac y B)-y=N$$ leads to Lambert function and the solution write $$y=B+(A+B)\,W\left(-\frac{B }{A+B}e^{-\frac{B+N}{A+B}}\right)$$

Claude Leibovici
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  • Hi @Claude, good answer. I think this might not cover all solutions though. For example, with $A=1$, $B=4$, we see that $F(y)$ can take all values in $\mathbb{R}^+$ for $-4<y<0$ but also, for $y>0$. So, for any $N>4$, we'd have $2$ solutions; one with $y<B$, as your answer states, but also one with $y>B$. An example of this is $A=1, B=4, N=5$, where we not only have $y\approx-3.786$ but also $y\approx8.392$. – Jam Sep 29 '17 at 09:54
  • @Jam. You are perfectly correct but I did not want to address the problem with $y\geq B$. Moreover, there may be multiples branches of $W(.)$ involved in the story. Why don't you provide some graphical illustrations ? Being almost blind, I cannot do it. Cheers. – Claude Leibovici Sep 29 '17 at 10:00
  • I was wondering about branches of $W$ but given that the limitation of $y<B$ comes in before we use $W$, I'm not certain that'd help. This is the graph I was using to understand the problem: https://i.imgur.com/QBPdBoT.png . The orange curve is $F(y)$ for $A=1, B=4$. The dashed vertical purple line is $y=B=4$ and the horizontal green line is $N=5$. So, since the horizontal line intersects $F(y)$ either side of the vertical line, it would suggest that there are $2$ solutions. I'm not sure how you'd find the other one, I'm partly just pointing it out for anyone reading this post :) – Jam Sep 29 '17 at 10:11
  • @Jam. Don't consider the solution. Select $y$ in the range $0\leq y\le 4$ and compute $n$. Plot $y$ as a function of $n$. It looks like an hyperbola. – Claude Leibovici Sep 29 '17 at 10:29
  • @Jam. It seems that you used $B+t$ instead of $B-t$. – Claude Leibovici Sep 29 '17 at 10:36