Computing $\int_0^\infty \frac{x^\lambda}{x+1} dx$

Question

I need help with integrating the following: $$\int_0^\infty \frac{x^\lambda}{x+1} \;dx \qquad \text{for }-1<\lambda<0$$

There is also a hint to place the branch cut of the integrand along the positive real axis. I'm not entirely sure how to use this hint, because if we place the branch cut along the real axis, then I can't perform contour integration involving the real axis...

I know at least this much:

• There is a pole at $z=-1$.
• The branch cut should start from the origin, and stretch out along the real axis to $+\infty$.

I suspect that I should be taking a path, just above the real axis, encircling the branch point at $z=0$, and perhaps a path back along just below the real axis. But once, again, I am not entirely sure whether that yields me the integral of interest. Any help to get me started will be appreciated.

Answer 1

When you have integrals of that form, you have a branch point and you have to use the well known Residues formula with the Cauchy Principal Value:

$$\mathcal{P} \int_0^{+\infty} z^{\lambda}R(z)\ \text{d}z = \frac{2\pi i}{1 - e^{2i\lambda \pi}} \sum_k\ \text{Res}[f(z), z_k]$$

Where $f(z) = z^{\lambda}R(z)$ and $R(z) = \frac{P(z)}{Q(z)}$ a rational function with certain conditions.

In your case, there is a simple pole at $z = -1$, hence

$$\text{Res} = \lim_{z\to 1} (z-1)f(z) = (-1)^\lambda = e^{i\pi\lambda}$$

hence

$$I = \frac{2\pi i}{1 - e^{2i\lambda \pi}} e^{i\pi\lambda}$$

Manipulate a bit

$$\frac{2\pi i}{-2i\sin(\lambda\pi)} = -\frac{\pi}{\sin(\lambda\pi)}$$

And with easy trigonometry, you can write it as

$$-\pi\ \text{cosec}(\lambda\pi)$$