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I need help calculating this integral:

$$\int_0^x \frac{2(e^{\gamma u}-1)}{(\gamma+\kappa)(e^{\gamma u}-1)+2\gamma} du$$

I tried with the integration by parts but the situation seems to get worse.

I'm not a skillful mathematician (not even a mathematician) so any help would be appreciated. I just know it's possible to solve it since it pops out in a famous financial model but haven't found any derivation yet.

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Dividing the numerator and denominator of the integral by $\gamma + \kappa$ gives $$\int_{0}^{x}\frac{a(e^{\gamma u}-1)du}{e^{\gamma u}-1+a\gamma}$$ Where $a=\frac2{\gamma + \kappa}.$

Breaking this into $2$ integrals, $$=\frac a\gamma\int_{0}^x\frac{\gamma e^{\gamma u}du}{e^{\gamma u}-1+a\gamma}-a\int_0^x\frac{du}{e^{\gamma u}-1+a\gamma}$$

For the first, substitute $e^{\gamma u}-1=t$ and you're left with $$\frac{a}{\gamma}\int \frac{dt}{t+a\gamma}$$ Do the same substitution for the second one, and you have $$\frac{1}{\gamma}\int \frac{dt}{(t+1)(t+a\gamma)}$$ The first one is now standard, and the second can be done by partial fractions.

Edit:

For the second integral, partial fractions isn't mandatory.

The second integral can be written as $$\frac{1}{\gamma(a\gamma-1)}\int \frac{(t+a\gamma)-(t+1)dt}{(t+1)(t+a\gamma)}$$ $$=\frac{1}{\gamma(a\gamma-1)}\int \frac{dt}{t+1}-\frac{1}{\gamma(a\gamma-1)}\int \frac{dt}{t+a\gamma}$$ Both of which are standard.

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  • $\begingroup$ First of all, thanks for the answer. I'm going to carefully read what you wrote. However, I ask you: is this partial fractions method easy to face for, as I wrote, a non-math student? If not, can you go through it too? $\endgroup$ – davidpaich May 23 '16 at 17:37
  • $\begingroup$ @davidpaich it's quite simple and straightforward after a few examples. Here's a quick and simple rundown: purplemath.com/modules/partfrac.htm $\endgroup$ – user300011 May 23 '16 at 17:39
  • $\begingroup$ @davidpaich see the edited answer $\endgroup$ – Nikunj May 23 '16 at 19:08
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Putting $\beta=\frac{2}{\gamma+\kappa}$ and $\alpha=\beta\gamma$ we have \begin{align} I&=\int_0^x \frac{2(\mathrm e^{\gamma u}-1)}{(\gamma+\kappa)(\mathrm e^{\gamma u}-1)+2\gamma} \mathrm du=\beta\int_0^x \frac{(\mathrm e^{\gamma u}-1)}{(\mathrm e^{\gamma u}-1)+\alpha} \mathrm du\\&=\beta\int_0^x \frac{(\mathrm e^{\gamma u}-1)\color{blue}{+\alpha-\alpha}}{(\mathrm e^{\gamma u}-1)+\alpha} \mathrm du=\beta x-\alpha\beta\int_0^x \frac{1}{(\mathrm e^{\gamma u}-1)+\alpha} \mathrm du\\ &=\beta x-\alpha\beta J \end{align} Putting $\delta=\alpha-1$ and with the substitution $\mathrm e^{\gamma u}=z$, $$ J=\int_0^x \frac{1}{(\mathrm e^{\gamma u}-1)+\alpha} \mathrm du=\int_0^{\mathrm e^{\gamma x}} \frac{1}{z+\delta} \frac{\mathrm dz}{\gamma z}=\frac{1}{\delta\gamma} \int_0^{\mathrm e^{\gamma x}} \left[ \frac{1}{z}-\frac{1}{z+\delta}\right]\mathrm dz=\frac{1}{\delta\gamma}\Big[\log z-\log(z+\delta)\Big]_0^{\mathrm e^{\gamma x}}=\frac{x}{\delta}-\frac{1}{\delta\gamma}\log\left(\mathrm e^{\gamma x}+\delta\right) $$ Finally $$ I=\beta x-\alpha\beta\left[\frac{x}{\delta}-\frac{1}{\delta\gamma}\log\left(\mathrm e^{\gamma x}+\delta\right)\right]=\frac{\beta}{1-\beta\gamma} \Big[x+\beta \log\left(\mathrm e^{\gamma x}+\beta\gamma-1\right)\Big] $$

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