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I input 3 functions as following,

g[n_, r_]=(2*Z^(3/2)*Hypergeometric1F1[1 - n, 2, (2*r*Z)/n])/(E^((r*Z)/n)*n^(3/2));
g1[n_, r_]=g[n,r]/.Z->m Z;
g2[n_, r_]=g[n,r]/.Z->m Z \[Alpha];

How ever the following three integral results shows Integral Results

Third Integral

Integrate[g2[n, r] g2[1, r] (m Z \[Alpha])/r r^2, {r, 0, \[Infinity]},
     Assumptions -> 
     n \[Element] Integers && n > 1 && m > 0 && Z > 0 && \[Alpha] > 0]

MMA says that the third integral does not converge. But since $mZ\alpha$ always product together, I expected that the third integral should just be the first integral with $Z\to mZ\alpha$. How to understand such results?

My MMA version number is 11.0.0.0 and the platform is Linux x86(64-bit).

Turgon
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superfishyrr
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  • Mathematica 11.3.0 on Windows outputs $$\frac{4 a m^2 \sqrt{n} \left(\frac{1-n}{n+1}\right)^n Z^2 (-\alpha )^n \alpha ^{1-n}}{n^2-1} $$ for the third integral. – user64494 Nov 19 '18 at 08:43
  • @user64494 Sadly my mma 11.3.0 outputs divergence as superfishyrr did, in both Linux and Windows version. Did you type something wrong? – Turgon Nov 20 '18 at 06:21
  • You are right. I replaced $\alpha$ by $a$, but only in the integral, not in $g2$. – user64494 Nov 20 '18 at 12:11
  • I can't explain why MMA's algorithm can't address this. But I can offer this: Your finding is not surprising. On this site, you'll find many examples of posters asking for help with challenging integrals that MMA can't solve, where the working solution is to simplify the integral into an equivalent form that MMA can. Here you've done the opposite; you've taken a challenging integral MMA can solve, and you've successively increased the complexity of the conditions to the point where the integral is in an equivalent form that MMA can't solve – theorist Nov 20 '18 at 21:46
  • perhaps related: https://mathematica.stackexchange.com/questions/23080/why-does-integrate-declare-a-convergent-integral-divergent – theorist Nov 20 '18 at 22:38

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