# Proving the equivalence of a sum and a double integral

Based on "Certain Subclass of Starlike Functions" journal by Chun-Yi and Shi-Qiong Zhou in 2007 (Science Direct), I found difficulties to understand the proof in Theorem 3 where they have verified:

$$1+2(1-\beta) \displaystyle\sum\limits_{n=2}^\infty \frac{z^{n-1}}{n(\alpha(n-1)+1)} = 1 +\frac{2(1-\beta)}{\alpha} \int_0^1 \! t^{\frac{1}{\alpha}} \,\int_0^1 \! \frac{vz}{1-tvz} \, \mathrm{d} v \ \mathrm{d} t$$

Could someone give me the idea of how to prove it?

Thank you.

-
Did you try expanding the innermost integrand as a geometric series, and swapping summation and integration? – J. M. Dec 7 '11 at 13:30
is there any bound on $|z|$? seems very similar to $\sum z^n = \frac{1}{1-z}$ – Ilya Dec 7 '11 at 13:30
@J.M.: I've tried but did not get the intended results. – Nina Dec 7 '11 at 13:34
@Ilya: they were inform that $|z|\leq r \rightarrow 1$ – Nina Dec 7 '11 at 13:41

Your ultimate goal is to prove that

$$\int\limits_0^1 {{t^{1/\alpha }}\int\limits_0^1 {\frac{{vz}}{{1 - tvz}}dv} dt} = \frac{1}{\alpha }\sum\limits_{n = 2}^\infty {\frac{{{z^{n - 1}}}}{{n\left( {\alpha \left( {n - 1} \right) + 1} \right)}}}$$

This can be done by geometric series since we're integrating over $(0,1$).

$$\frac{{vz}}{{1 - tvz}} = vz\sum\limits_{k = 0}^\infty {{t^k}{v^k}{z^k}} = \sum\limits_{k = 0}^\infty {{t^k}{v^{k + 1}}{z^{k + 1}}}$$

Thus we have

$$\int\limits_0^1 {\frac{{vz}}{{1 - tvz}}} dv = \sum\limits_{k = 0}^\infty {\frac{{{t^k}{z^{k + 1}}}}{{k + 2}}}$$

Moving on we get:

$$\int\limits_0^1 {\sum\limits_{k = 0}^\infty {\frac{{{t^{k + 1/\alpha }}{z^{k + 1}}}}{{k + 2}}} } dt = \sum\limits_{k = 0}^\infty {\frac{{{z^{k + 1}}}}{{\left( {k + 2} \right)\left( {k + 1 + 1/\alpha } \right)}}}$$

Rearranging to $k=2$ we get

\eqalign{ & \sum\limits_{k = 2}^\infty {\frac{{{z^{k - 2 + 1}}}}{{\left( {k - 2 + 2} \right)\left( {k - 2 + 1 + 1/\alpha } \right)}}} \cr & \sum\limits_{k = 2}^\infty {\frac{{{z^{k - 1}}}}{{k\left( {k - 1 + 1/\alpha } \right)}}} \cr & \frac{1}{\alpha }\sum\limits_{k = 2}^\infty {\frac{{{z^{k - 1}}}}{{k\left( {\alpha \left( {k - 1} \right) + 1} \right)}}} \cr}

which is what you wanted.

-

Subtract $1$, then multiply by $\frac{\alpha}{2(1-\beta)}$ on both sides. Expand (using a sum from $n=2$ to $\infty$ as in the desired result): $$\frac{vz}{1-tvz}=\sum_{n=2}^{\infty}t^{n-2}(vz)^{n-1}$$ Exchange summation and inside integration (inside the radius of convergence, where the series is absolutely convergent), and use the fact that $\int_0^1v^kdv=\frac{1}{k+1}$ (for $k>-1$), then add the exponents of $t$, swap summation and integration again and use the same fact.

-
You need to assume $|tvz|<1$. – Paul Dec 7 '11 at 14:02
@Paul: Nina gave a condition like that in the comments... – J. M. Dec 7 '11 at 14:04
Your exponents in the sum should be $k$ not $n$ – deinst Dec 7 '11 at 14:06
is it anything to do with lerch zeta function? From my research, I can see the possibility but I can't confirm yet. – Nina Dec 7 '11 at 14:13
@J.M.: I have tried to make use that application, since in the denominator have $n^{2}$, I already use partial fraction to separate them.. I only obtain this result: $\int_0^1 \! \frac{w^{\frac{1}{\alpha}}}{1-zw} \mathrm{d} w - \alpha \int_0^1 \! \frac{x}{1-zx}\mathrm{d} x$ – Nina Dec 8 '11 at 2:16