Factorial series(!)

$\displaystyle \large \int_{0}^{c} \frac{sinax}{bx}dx$ is

$\displaystyle \large \frac{1}{b} \int_{0}^{ac} \frac{sint}{t}dt$ by putting $ax=t$

now making use of expansion for $sinx$ ,

$\displaystyle \large \frac{1}{b} \int_{0}^{ac} (1-\frac{t^{2}}{3!}+\frac{t^{4}}{5!}-...)dt$

this is $\displaystyle \large \frac{1}{b} [ac-\frac{(ac)^{3}}{4!-3!}+\frac{(ac)^{5}}{6!-5!}-...]$

the summation asked is $\displaystyle \large \sum_{1}^{\infty} \frac{(-1)^{n-1}2^{2n-1}}{2(2n-1)(2n-1)!}$

clearly, $\large ac=2 ; b=2$

$\large {a+b+c=5}$

Note that:

$\begin{aligned} S & = \frac 1{2!-1!} - \frac 4{4!-3!} + \frac {16}{6!-5!} - \frac {64}{8!-7!} + ... \\ & = \frac {2^0}{(2-1)1!} - \frac {2^2}{(4-1)3!} + \frac {2^4}{(6-1)5!} - \frac {2^6}{(8-1)7!} + ... \\ & = \frac {2^0}{1 \cdot 1!} - \frac {2^2}{3 \cdot 3!} + \frac {2^4}{5 \cdot 5!} - \frac {2^6}{7 \cdot 7!} + ... \end{aligned}$

Now consider:

$\begin{aligned} \sin x & = \frac x{1!} - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + ... \\ \frac {\sin x}x & = \frac 1{1!} - \frac {x^2}{3!} + \frac {x^4}{5!} - \frac {x^6}{7!} + ... \\ \int \frac {\sin x}x dx & = \frac x{1 \cdot 1!} - \frac {x^3}{3 \cdot 3!} + \frac {x^5}{5 \cdot 5!} - \frac {x^7}{7 \cdot 7!} + ... \\ \int_0^2 \frac {\sin x}x dx & = \frac 2{1 \cdot 1!} - \frac {2^3}{3 \cdot 3!} + \frac {2^5}{5 \cdot 5!} - \frac {2^7}{7 \cdot 7!} + ... \\ \int_0^2 \frac {\sin x}{2x} dx & = \frac {2^0}{1 \cdot 1!} - \frac {2^2}{3 \cdot 3!} + \frac {2^4}{5 \cdot 5!} - \frac {2^6}{7 \cdot 7!} + ... = S \end{aligned}$

$\implies a+b+c = 1+2+2 = \boxed{5}$

By Maclaurin's series we have

sinx = x-x^3/3!+x^5/5!-x^7/7!

As we can see in the denominator of our series we have terms like 1 1! , 2 2!

So quite naturally we divide by x (so that on integration we get that)

sinx/x = 1-x^2/3!+x^4/5! ......

On integrating this we wont be able to get the required series . but if we replace x by 2x then

sin(2x)/2x = 1-4x^2/3!+16x^4/5!

Now just integrate this from 0 to 1 to get the required series

a = 2 b= 2 and c=1