samples/texts/1085779/page_14.md

Theorem 4.

$${\mathcal{H}}^{q+1}({\left(\frac{1}{a}\right)}_{r(2)};1_{r(2)};\beta +2)=a^{\beta +2}({\mathrm{L}\mathrm{i}}_q\left(\frac{1}{a}\right)-(\beta +1){\mathrm{L}\mathrm{i}}_{q+1}\left(\frac{1}{a}\right)+(\beta +1)\sum _{k=1}^{\beta }\frac{1}{a^k{k}^{q+1}}-\sum _{k=1}^{\beta }\frac{1}{a^k{k}^q})\backslash mathcal\{H\}^\{q+1\}\; \backslash left(\; \backslash left(\backslash frac\{1\}\{a\}\backslash right)\_\{r(2)\}\; ;\; 1\_\{r(2)\};\; \backslash beta\; +\; 2\; \backslash right)\; =\; a^\{\backslash beta+2\}\; \backslash left(\; \backslash mathrm\{Li\}\_q\; \backslash left(\backslash frac\{1\}\{a\}\backslash right)\; -\; (\backslash beta+1)\; \backslash mathrm\{Li\}\_\{q+1\}\; \backslash left(\backslash frac\{1\}\{a\}\backslash right)\; +\; (\backslash beta+1)\; \backslash sum\_\{k=1\}^\{\backslash beta\}\; \backslash frac\{1\}\{a^k\; k^\{q+1\}\}\; -\; \backslash sum\_\{k=1\}^\{\backslash beta\}\; \backslash frac\{1\}\{a^k\; k^q\}\; \backslash right)$$

Proof. As per the definition of Hyder Series

$${\mathcal{H}}^{q+1}({\left(\frac{1}{a}\right)}_{r(2)};1_{r(2)};\beta +2)=\sum _{k_1=0}^{\mathrm{\infty }}\sum _{k_2=0}^{\mathrm{\infty }}\frac{1}{a^{k_1+k_2}(k_1+k_2+\beta +2{)}^{q+1}}\backslash mathcal\{H\}^\{q+1\}\; \backslash left(\; \backslash left(\; \backslash frac\{1\}\{a\}\; \backslash right)\_\{r(2)\}\; ;\; 1\_\{r(2)\}\; ;\; \backslash beta\; +\; 2\; \backslash right)\; =\; \backslash sum\_\{k\_1=0\}^\{\backslash infty\}\; \backslash sum\_\{k\_2=0\}^\{\backslash infty\}\; \backslash frac\{1\}\{a^\{k\_1+k\_2\}(k\_1+k\_2+\backslash beta+2)^\{q+1\}\}$$

To evaluate the series, we will make use of the Eq(11), In this case $m = k_1 + k_2 + \beta + 1$

therefore

The above mentioned Integral A is the same Integral that we just came to prove in Lemma 2. Therefore the final result becomes

$${\mathcal{H}}^{q+1}({\left(\frac{1}{a}\right)}_{r(2)};1_{r(2)};\beta +2)=a^{\beta +2}({\mathrm{Li}}_q\left(\frac{1}{a}\right)-(\beta +1){\mathrm{Li}}_{q+1}\left(\frac{1}{a}\right)+(\beta +1)\sum _{k=1}^{\beta }\frac{1}{a^k{k}^{q+1}}-\sum _{k=1}^{\beta }\frac{1}{a^k{k}^q})\backslash mathcal\{H\}^\{q+1\}\; \backslash left(\; \backslash left(\backslash frac\{1\}\{a\}\backslash right)\_\{r(2)\}\; ;\; 1\_\{r(2)\}\; ;\; \backslash beta\; +\; 2\; \backslash right)\; =\; a^\{\backslash beta+2\}\; \backslash left(\; \backslash operatorname\{Li\}\_q\; \backslash left(\backslash frac\{1\}\{a\}\backslash right)\; -\; (\backslash beta+1)\; \backslash operatorname\{Li\}\_\{q+1\}\; \backslash left(\backslash frac\{1\}\{a\}\backslash right)\; +\; (\backslash beta+1)\; \backslash sum\_\{k=1\}^\{\backslash beta\}\; \backslash frac\{1\}\{a^k\; k^\{q+1\}\}\; -\; \backslash sum\_\{k=1\}^\{\backslash beta\}\; \backslash frac\{1\}\{a^k\; k^q\}\; \backslash right)$$