I have this sum $$\sum_{i=0}^{m} \exp [(\frac{a}{b+c+i})^2] $$ where the upper limit $m$ is a finite non-negative integer, and $a,b,c\in\mathbb{R}$. I want to transform summation to an integral using the Euler–Maclaurin formula, or any other method if possible. How can I do this?
P.S. My major is not Math; searching on the internet, I see different versions of the Euler–Maclaurin formula that I do not know which one applies to my case.
Let $b+c=d$. We know that $$\int e^{\frac{a}{(d+i)^2}}\,di=(d+i)\, e^{\frac{a}{(d+i)^2}}-\sqrt{a\pi } \, \text{erfi}\left(\frac{\sqrt{a}}{d+i}\right)$$
Using the simplest form of Euler-MacLaurin summation $$\sum_{i=0}^m e^{\frac{a}{(d+i)^2}}\sim m+C-\frac a m+O\left(\frac{1}{m^2}\right)$$
$$C=d+\frac{1}{2} \left(1+e^{\frac{a}{d^2}}\right)+ \sqrt{a \pi}\, \text{erfi}\left(\frac{\sqrt{a}}{d}\right)+e^{\frac{a}{d^2}}\, A$$ where $$A=-d+\frac{a}{6 d^3}-\frac{a}{30 d^5}+\frac{(10-21 a) a}{420 d^7}+\frac{a \left(-7 a^2+55 a-21\right)}{630 d^9}+\frac{a^2 (85 a-273)}{1260 d^{11}}+$$ $$\frac{a^3 (20 a-399)}{1260 d^{13}}+\frac{a^4 (16 a-2499)}{15120 d^{15}}-\frac{13 a^5}{360 d^{17}}-\frac{a^6}{300 d^{19}}-\frac{a^7}{9450 d^{21}}$$
Trying for $a=2$, $d=5$ and $m=100$, converted to decimals, the above gives $101.42993$ while the exact summation gives $101.43097$ (absolute error of $0.00104$).
Using the next level of summation would give $101.43102$ (absolute error of $0.00005$).
