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After 2 change of variables, and a $x = \log(u)$ transformation, I have this integral...

$$\int_0^{\infty } \frac{\alpha e^x}{\lambda-e^x \lambda+e^{\alpha x}} \, dx$$

What are some recommendations on how to integrate this?

NOTE: The original integral looked like:

$$\int_{1 }^{\infty } \frac{\alpha \left(\frac{1 }{x}\right)^{\alpha }}{1-\lambda (x-1 ) \left(\frac{1 }{x}\right)^{\alpha }} \, dx$$

and for the integral to converge $\alpha > 1$.

Finally, when $\alpha = 2$, I did get this expression...

$$\frac{8 \tan ^{-1}\left(\frac{\sqrt{\lambda }}{\sqrt{4-\lambda }}\right)}{\sqrt{(4-\lambda ) \lambda }}$$

Thanks,

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  • 2
    $\begingroup$ The substitution $u:=e^x$ turns the integral into $$\int_1^\infty\frac{\alpha}{\lambda-\lambda u + u^\alpha}du$$ which is still pretty bad but maybe more tractable. Do you know anything about $\alpha$? $\endgroup$ – Tom Feb 25 '17 at 13:16
  • $\begingroup$ Yes - for the integral to converge, $\alpha > 1$ $\endgroup$ – PiE Feb 25 '17 at 13:17
  • $\begingroup$ @Tom - I tried that substitution, but didn't make much progress...but then again...Next, I transformed it into the one on top... $\endgroup$ – PiE Feb 25 '17 at 13:20
  • $\begingroup$ to get a nice closed form you should at least have the transoformed numerator to be a polynomial in $x$ $\endgroup$ – tired Feb 25 '17 at 13:52
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For integer $\alpha>1$, I got the following:

Take the Mellin transform of the integrand with respect to $\lambda$, $$ \int_0^\infty \frac{\alpha e^x}{\lambda-e^x \lambda+e^{\alpha x}} \lambda^{s-1} \; d\lambda = \frac{\alpha e^{x-\alpha x}}{(-e^{-\alpha x}(e^x-1))^s}\pi\csc(\pi s) = I_{\alpha}(s,x) $$ now integrate over $x$ $$ \int_0^\infty I_{\alpha}(s,x) \; dx = (-1)^{-s}a\pi\csc(\pi s)\frac{\Gamma(1-s)\Gamma(\alpha-1+s-\alpha s)}{\Gamma(\alpha-\alpha s)} $$ now take the inverse Mellin transform with respect to $s$ to get something that depends on $\alpha$ and $\lambda$, $$ G_\alpha(\lambda) = \frac{1}{2 \pi i}\int_{c-i \infty}^{c + i \infty} \lambda^{-s}(-1)^{-s}a\pi\csc(\pi s)\frac{\Gamma(1-s)\Gamma(\alpha-1+s-\alpha s)}{\Gamma(\alpha-\alpha s)}\;ds $$ we can use the Mathematica command

InverseMellinTransform[((-1)^-s a [Pi] Csc[[Pi] s] Gamma[ 1 - s] Gamma[-1 + a + s - a s])/Gamma[a - a s] /. a -> 2, s, l]

Where the a -> X part can be changed for various values of $\alpha$. This reproduces your result for $\alpha=2$, but that obscures the hypergeometric nature of the result. It also gives results for higher values of $\alpha$ as hypergeometric functions weighted by products of gamma functions.

The results seem to have a pattern $$ G_4(\lambda)=\frac{3^2}{2^3}\sqrt{\frac{3}{2}}\frac{\Gamma(\frac{4}{3})\Gamma(\frac{5}{3})}{\Gamma(\frac{5}{4})\Gamma(\frac{7}{4})}\;_4F_3\left(\frac{3}{3},\frac{3}{3},\frac{4}{3},\frac{5}{3}\bigg|\frac{5}{4},\frac{6}{4},\frac{7}{4}\bigg|\frac{3^3}{4^4}\lambda\right) $$ $$ G_5(\lambda)=\pi\frac{4^3}{5^3}\sqrt{\frac{2}{5}}\frac{\Gamma(\frac{5}{4})\Gamma(\frac{7}{4})}{\Gamma(\frac{6}{5})\Gamma(\frac{7}{5})\Gamma(\frac{8}{5})\Gamma(\frac{9}{5})}\;_5F_4\left(\frac{4}{4},\frac{4}{4},\frac{5}{4},\frac{6}{4},\frac{7}{4}\bigg|\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5}\bigg|\frac{4^4}{5^5}\lambda\right) $$ In general it appears that $$ G_n=C_n\; _nF_{n-1}\left(\frac{n-1}{n-1},\frac{n-1}{n-1},\frac{n}{n-1},\cdots,\frac{2n-3}{n-1}\bigg|\frac{n+1}{n},\frac{n+2}{n},\cdots,\frac{2n-1}{n}\bigg|\frac{(n-1)^{n-1}\lambda}{n^n}\right) $$ where $$ C_n=\sqrt{\frac{2\pi(n-1)}{n}}\frac{(n-1)^{n-2}}{n^{n-2}}\frac{\prod_{k=0}^{n-2} \Gamma(\frac{n+k}{n-1})}{\prod_{k=1}^{n-1} \Gamma(\frac{n+k}{n})} $$ where we see your result in $C_2=2$ and $$ _2F_1\left(1,1\bigg|\frac{3}{2}\bigg|\frac{\lambda}{4}\right) = \frac{4 \sin^{-1}(\sqrt{\lambda}/2)}{\sqrt{(4-\lambda)\lambda}} $$ seems to check out for a few numerical examples.

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