# Is there an elementary method for evaluating $\int_0^\infty \frac{dx}{x^s (x+1)}$?

I found a way to evaluate $\int_0^\infty \frac{dx}{x^s (x+1)}$ using the assumption that $s\in\mathbb{R}$ and $0<s<1$.

Apparently it should be easily extended to all $s\in\mathbb{C}$ with $0<Re(s)<1$.

I posted my solution here: http://thetactician.net/Math/Analysis/Integral1.pdf

I'm pretty sure there's a more concise method for evaluating it...and I'd also like to make the extension to $\mathbb{C}$ more rigorous.

Any ideas?

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Knowledge on some special functions (especially Gamma function and Beta function) allows us to evaluate the integral in few lines. You can confirm at here. However, I'm not sure you will agree that this is elementary. –  sos440 Feb 29 '12 at 16:58

First, let $x=\frac{1}{t}$. Then our integral is $$\int_{0}^\infty \frac{t^{s-1}}{t+1}dt$$ which is the Mellin transform of the function $\frac{1}{1+t}$. In this math stack exchange answer it is shown that $$\mathcal{M}\left(\frac{1}{1+x^{b}}\right)(s)=\int_0^\infty \frac{t^{s-1}}{1+t^b}dt =\frac{\pi}{b}\csc\left(\frac{\pi s}{b}\right).$$

This solution uses the Beta function, and an identity relating it to the Gamma function which you may or may not consider to be elementary. (There are proofs of this identity which use no complex analysis)

In each of the following threads, there are answers which are of interest:

Simpler way to compute a definite integral without resorting to partial fractions?

$\int_{0}^{\infty}\frac{dx}{1+x^n}$

Closed form of integral.

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I actually arrived at this problem in trying to prove the reflection formulas for the Gamma and Pi functions...so it would amount to a circular proof...but yeah, interesting that you would arrive at what amounts to the reverse. Thanks for the suggestions, though. I'd be interested in the proofs you speak of that use no complex analysis... –  Eric Feb 29 '12 at 21:12

Not elementary (in the sense of not using complex analysis), but this is how I'd do it:

Let $$f(z) = \frac{1}{z^s(z+1)},$$ where $z^s$ denotes the "natural branch", i.e. choose $\phi \arg z \in (0,2\pi)$ and put $(re^{i\phi})^s = r^s e^{is\phi}$. Take a "keyhole contour" C:

and integrate $f$ along $C$ using the residue theorem:

$$\int_C f(z)\,dz = 2\pi i \operatorname{Res}_{z=-1} f(z).$$

Estimating $\int_\gamma f(z)\,dz$ and $\int_\Gamma f(z)\,dz$, we have

$$\left| \int_\gamma f(z)\,dz \right| \le \frac{M}{r^{\mathrm{Re}(s)}} \cdot 2\pi r \to 0 \qquad\text{as }r\to0$$ and $$\left| \int_\Gamma f(z)\,dz \right| \le \frac{M}{R^{1+\mathrm{Re}(s)}} \cdot 2\pi R \to 0 \qquad\text{as }R\to\infty.$$

(For the first estimate, we want $\mathrm{Re}(s) < 1$ and for the second $\mathrm{Re}(s) > 0$.

For the two remaining line segments, on the "top" segment we get the integral we're looking for as $r \to 0$ and $R\to\infty$. On the "bottom" segment, we get (remember the choice of branch) $$-\int_0^\infty \frac{1}{x^s e^{2\pi i s} (1+x)}\,dx.$$

Putting it all together: $$(1-e^{-2\pi i s}) \int_0^\infty \frac{1}{x^s(1+x)}\,dx = 2\pi i \operatorname{Res}_{z=-1} f(z) = 2\pi i (-1)^{-s} = 2\pi i e^{-\pi i s}$$ so $$\int_0^\infty \frac{1}{x^s(1+x)}\,dx = 2\pi i \frac{e^{-\pi i s}}{1-e^{-2\pi i s}} = \frac{\pi}{\sin s\pi}.$$

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Thank you. I thought there probably was a way to apply the residue theorem but I wasn't sure what contour to use. –  Eric Mar 1 '12 at 0:52
Worth linking to: math.stackexchange.com/a/110465/6075 –  Eric Naslund Mar 1 '12 at 7:58
It would be interesting to hear the reasons for a sudden downvote almost a year after this answer was posted. –  mrf Feb 20 '13 at 13:22