$\small \int_0^\infty \frac{\sin(x)}{e^x-1} dx$

To solve the integral, we use the series representation of $\frac1{1-x}$ , $\frac1{1-x}=1+x+x^2+x^3+\dots$ We can't directly substitute $e^x$ in, because the series diverges for $x>1$ . Instead, we can rearrange and use $e^{-x}$ , which is between 0 and 1 for all positive $x$ . $\begin{aligned} \int_0^\infty \frac{\sin x}{e^x-1} dx &=\int_0^\infty \frac{e^{-x}\sin x}{1-e^{-x}} dx\\ &=\int_0^\infty e^{-x}\sin x \sum_{n=0}^\infty \left(e^{-x}\right)^n dx\\ &=\int_0^\infty \sum_{n=0}^\infty e^{-xn-x}\sin x\ dx\\ &=\int_0^\infty \sum_{n=1}^\infty e^{-xn}\sin x\ dx \end{aligned}$ Now, consider the integral of a single element of this sum, which we can solve with integration by parts: $\begin{aligned} I &=\int_0^\infty e^{-xn}\sin x\ dx\\ &=\left[-e^{-xn}\cos x\right]_0^\infty-n\int_0^\infty e^{-xn}\cos x\ dx\\ &=1-n\left(\left[e^{-xn}\sin x\right]_0^\infty+n\int_0^\infty e^{-xn}\sin x\ dx\right)\\ &=1-n\left(0+nI\right)\\ &=1-n^2I\\ \implies I(1+n^2)&=1\\ \implies I&=\frac1{1+n^2}\end{aligned}$ Now, since the integral exists for any $n\geq1$ , we can swap the integral and summation: $\begin{aligned} \int_0^\infty \frac{\sin x}{e^x-1} dx &=\int_0^\infty \sum_{n=1}^\infty e^{-xn}\sin x\ dx\\ &=\sum_{n=1}^\infty \int_0^\infty e^{-xn}\sin x\ dx\\ &=\sum_{n=1}^\infty \frac1{1+n^2}\end{aligned}$ We can solve this using the digamma function $\psi(x)$ , which has the following property: $\begin{aligned}\psi(z+1) &= -\gamma+\sum_{k=1}^\infty \left(\frac1k-\frac1{k+z}\right)\\ \implies\psi(1+z)-\psi(1-z) &= -\gamma+\sum_{k=1}^\infty \left(\frac1k-\frac1{k+z}\right)+\gamma+\sum_{k=1}^\infty \left(-\frac1k+\frac1{k-z}\right)\\ &= \sum_{k=1}^\infty \left(\frac1{k-z}-\frac1{k+z}\right)\end{aligned}$ Now, we can use complex numbers to write the sum in this form: $\begin{aligned}\sum_{n=1}^\infty \frac1{1+n^2} &=\sum_{n=1}^\infty \frac1{(n+i)(n-i)}\\ &=\sum_{n=1}^\infty \frac1{2i}\left(\frac1{n-i}-\frac1{n+i}\right)\\ &=-\frac i2\left(\psi(1+i)-\psi(1-i)\right)\end{aligned}$ $\psi(1+i)$ can be simplified using the digamma recurrence formula, $\psi(x+1)=\psi(x)+\frac1x$ .

$\psi(1-i)$ can be simplified using the digamma reflection formula, $\psi(1-x)=\psi(x)+\pi\cot\pi x$ . Hence, we have $\begin{aligned}\sum_{n=1}^\infty &=-\frac i2\left(\psi(1+i)-\psi(1-i)\right)\\ &=-\frac i2\left(\psi(i)+\frac1i-\psi(i)-\pi\cot\pi i\right)\\ &=-\frac i2\left(\frac1i-\pi\cot\pi i\right)\\ &=-\frac12+\frac{\pi}2\cot\pi i\end{aligned}$ We now use the complex cotangent formula (which can be derived from Euler's formula $e^{ix}=\cos x+i\sin x$ ): $\begin{aligned}-\frac12+\frac{\pi i}2\cot\pi i &=-\frac12+\frac{\pi i}2 i\left(\frac{e^{i(\pi i)}+e^{-i(\pi i)}}{e^{i(\pi i)}-e^{-i(\pi i)}}\right)\\ &=-\frac12-\frac{\pi}2\left(\frac{e^{-\pi}+e^{\pi}}{e^{-\pi}-e^{\pi}}\right)\\ &=-\frac12-\frac{\pi}2\left(\frac{1+e^{2\pi}}{1-e^{2\pi}}\right)\\ &=-\frac12-\frac{\pi}2\left(\frac2{1-e^{2\pi}}-1\right)\\ &=-\frac12+\frac{\pi}2-\frac{\pi}{1-e^{2\pi}}\\ &=\frac{\pi-1}2+\frac{\pi}{e^{2\pi}-1}\end{aligned}$ So $\displaystyle\int_0^\infty \frac{\sin x}{e^x-1}dx=\frac{\pi-1}2+\frac{\pi}{e^{2\pi}-1}$ , and hence $a=1,b=2,c=2$ and so the answer is $\boxed{5}$

Someone can probably write a better solution but heres how I got it.

So starting over at wolfram
I found that the integral evaluates to 1.07667

I then had to write a quick brute force to check positive integer values for a, b and c until we got a combo that matched.
I had to "match" the beginning of the string to account for decimals and rounding errors and whatnot.

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<?php
for($a=1;$a<10;$a++){
    for($b=1;$b<10;$b++){
        for($c=1;$c<10;$c++){
            $v = ((M_PI-$a)/$b)+(M_PI/(pow(M_E,$c*M_PI)-1));
            if(stristr($v,'1.07667')){
                echo 'f(abc)='.$v.' '.$a.'+'.$b.'+'.$c.'='.($a+$b+$c);exit;
            }
        }
    }
}
// prints: f(abc)=1.0766740474686 1+2+2=5
?>