i need help with the following integral please:
v.p $\int_{-\infty}^{\infty} \frac{e^{px}}{1-e^{x}}\, dx$
$0<p<1$
Where v.p denotes the main value.
I have tried to do it through excercises that i have seen similars, but using the main value i have seen very few like this.
Thank you very much in advance.
Use the contour below where $T \to \infty$, and $\varepsilon \to 0$ to avoid the singularities at $0$ and $2\pi i$.
$$I=\int\limits_{C_1} f(z) \; dz+\int\limits_{C_3} f(z) \; dz$$which can be determined by $z=x$.
By Cauchy's first integral theorem: $$I=-\int\limits_{C_2}-\int\limits_{C_4}-\int\limits_{C_5}-\int\limits_{C_6}-\int\limits_{C_7}-\int\limits_{C_8}$$
For $C_4$ let $z=T+iy$ with $0 \leq y \leq 2\pi$, and similarly for $C_8$ let $z=-T+iy$ with $2\pi \leq y \leq 0$.
You'll see that $\int\limits_{C_8} f(z) \; dz$ and $\int\limits_{C_4} f(z) \; dz$ go to zero when $0<p<1$.
$$I=-\int\limits_{C_2}-\int\limits_{C_5}-\int\limits_{C_6}-\int\limits_{C_7}$$
For $C_5$ and $C_7$ let $z=x+2\pi i$, and $T \leq x \leq \varepsilon$ and $-\varepsilon \leq x \leq -T$, respectively: $$\int_T^{\varepsilon} \frac{e^{p(x+2\pi i)}}{1-e^{x+2\pi i}} \; dx+\int_{-\varepsilon}^{-T} \frac{e^{p(x+2\pi i)}}{1-e^{x+2\pi i}} \; dx=I(-e^{2\pi p i})$$
and so we are left with: $$\left(1-e^{2\pi p i}\right)I=-\int\limits_{C_2}-\int\limits_{C_6}$$
For $C_2$ let $z=\varepsilon e^{i \theta}$ with $\pi \leq \theta \leq 0$:
$$\int\limits_{C_2}=\int_{\pi}^0 \frac{e^{p\varepsilon e^{i \theta}}}{1-e^{\varepsilon e^{i \theta}}}i\varepsilon e^{i \theta} \; d\theta$$
With $\lim_{\varepsilon \to 0}$ $$\int\limits_{C_2}=i \int_0^{\pi} d\theta=\pi i$$
For $C_6$ let $z=2\pi i+\varepsilon e^{i \theta}$ with $0 \leq \theta \leq -\pi$, and with the same process as solving for $C_2$, the integral simplifies to:
$$C_6=-ie^{p2\pi i}\int_0^{-\pi} d\theta=i\pi e^{p2\pi i}$$
Now, $$\left(1-e^{2\pi p i}\right)I=-\left(\pi i+i\pi e^{p2\pi i}\right)$$ $$I=\pi i \frac{2\cos{(p\pi)}}{2i\sin{(p \pi)}}$$ $$I=\boxed{\frac{\pi}{\tan{\left(p \pi\right)}}}$$