200 Followers Problem

Well, nice problem!

$f(x) = {log}_{2}e - {log}_{4}e + {log}_{6}e...$

We know that ${log}_{a}b = \frac{ln(b)}{ln(a)}$

Hence, $f(x)$ becomes -

$\displaystyle \sum_{k=1}^{\infty}{\frac{{(-1)}^{k-1}}{ln(2k)}}$

Now, we know that $\displaystyle \int{{a}^{x}} = \frac{{a}^{x}}{ln(a)}$

Io, we have got the natural logarithm in the denominator!

We here would try to guess about the bounds. As we can see, we need $1$ in the numerator and that ${a}^{0} = 1$ . Also, the 2nd bound must make the function zero. Well... what's $ln(0)$ ? You got it right - $-\infty$ , so we have got the other too.

$\displaystyle \frac{1}{ln(2k)} = \int_{-\infty}^{0}{{(2k)}^{x}}$

Well, we can do 2 more things,

$\displaystyle \frac{1}{ln(2k)} = (-1)(-1)\int_{0}^{-\infty}{{(2k)}^{-x}}$

Put in the sum,

$\displaystyle \sum_{k=1}^{\infty}{{(-1)}^{k-1}\int_{0}^{-\infty}{{(2k)}^{-x}}}$

Getting integral sign outside,

$\displaystyle \int_{0}^{-\infty}{\sum_{k=1}^{\infty}{\frac{{(-1)}^{k-1}}{{(2k)}^{x}}}}$

$\displaystyle \int_{0}^{-\infty}{{2}^{-x}\sum_{k=1}^{\infty}{\frac{{(-1)}^{k-1}}{{k}^{x}}}}$

Now, we will first find out the sum -

We know $\displaystyle \zeta(x) = \sum_{k=1}^{\infty}{\frac{1}{{a}^{x}}}$

Multiplying ${2}^{1-x}$ both sides,

$\displaystyle ({2}^{1-x}\zeta(x) = \sum_{k=1}^{\infty}{\frac{2}{{2a}^{x}}}$

Subtracting these 2 equations(1-2),

$\displaystyle {2}^{-x}({2}^x - 2)\zeta(x) = \sum_{k=1}^{\infty}{\frac{{(-1)}^{k-1}}{{a}^{x}}}$

Hence, the function under integral sign equates to

$\displaystyle {2}^{-2x}({2}^{x} - 2)\zeta(x) = {2}^{-x}\zeta(x) - {2}^{1-2x}\zeta(x)$

As a result,

$\displaystyle \int_{0}^{-\infty}{{2}^{-x}\zeta(x) - {2}^{1-2x}\zeta(x)}$