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I have been trying to solve a simple mathematical integral. I know that the solution exists but it is not being verified by Matlab. Here is the integral that I am solving along with its answer which I found from "Table of Integrals" Eq. 3.351.1. $1- \dfrac{2}{\mathcal{R}^2_{\mathcal{G}}}\int_{0}^{\mathcal{R}_{\mathcal{G}}} \exp (-\Phi r^{\alpha}) r dr,$ where $\alpha>0, \mathcal{R}_\mathcal{G}>0$. I substituted $x\rightarrow r^\alpha$ and applied 3.351.1 from the above-mentioned book. The answer that I found is $(\Phi)^{\Bigg((\dfrac{2-\alpha}{\alpha})-1\Bigg)} \gamma\Bigg((\dfrac{2-\alpha}{\alpha})+1),(\Phi {\mathcal{R}_\mathcal{G}}^\alpha)\Bigg)$.

When I try to compare the answers in Matlab, they don't match. I even tried to approximate the integral with Gauss-Chebyshev Quadrature but the integral answer obtained doesn't match with the approximated answer. Its weird. Any idea on this? Thanks

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I do not see the problem except that you are missing a factor.

Consider $$I=\int e^{-\Phi\, r^{\alpha}}\, r\, dr$$ let, as you did $$r^{\alpha}=x \implies r=x^{\frac{1}{\alpha}}\implies dr=\color{red}{\frac{1}{\alpha}}x^{\frac{1}{\alpha}-1}\implies I=\color{red}{\frac{1}{\alpha}}\int e^{-\Phi x}\,x^{\frac{2}{\alpha}-1}\,dx$$ Now, apply the formula given in the book with $n=\frac{2}{\alpha}-1$ and $\mu=\Phi$. I prefer it in the form $$\int_0^u x^n\, e^{-\mu x}\,dx=\mu ^{-(n+1)} \,(\Gamma (n+1)-\Gamma (n+1, \mu u ))$$

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  • $\begingroup$ Thank you so much, I included $\dfrac{1}{\alpha}$ in my solution on paper but missed to mention here. The above formula is valid if n is positive. How do we proceed if value of n is negative? I am really thankful for your input $\endgroup$ – hakkunamattata Sep 14 '18 at 6:54

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