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I need to prove that

$$\int \limits_{0}^{\infty} \frac{x^{p-1}}{1+x} dx = \frac{\pi}{\sin (p \pi)}$$ for $0<p<1$.

I know how to do it using residue theorem and using the Beta function integral definitions. Also, I can transform the integral to a summation; but that is cheating.

So, is there any simpler way to evaluate the integral using the integral definition of gamma function, or something lighter than the residue theorem?

Quanto
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Ahmad
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    I don't know, maybe Laplace transform? According to Mathematica the Laplace transform is somewhat simple: $$\mathcal{L}\left(x\mapsto \frac{x^{p-1}}{1+x}\right)(s)=e^s\Gamma(p)\Gamma(1-p,s)$$ Though the incomplete Gamma function does appear. – K.defaoite May 09 '21 at 21:51
  • @K.defaoite thanks but not simple enough, at least with the beta function i have a chance. – Ahmad May 09 '21 at 21:57
  • See the methods in the question posted HERE. – Mark Viola May 09 '21 at 22:04
  • The reference to a previously answered question requires the substitution $x = u^2$ to form a complete answer to this question. – Eric Towers May 09 '21 at 22:05
  • @MarkViola so no other way than Beta function – Ahmad May 09 '21 at 22:16
  • @Ahmad : Is there a reason you are ignoring the referenced prior question and its answers? – Eric Towers May 09 '21 at 22:16
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    @EricTowers not really but i want the simpler way if possible – Ahmad May 09 '21 at 22:29
  • The referenced question's answers is nothing but simpler ways. – Eric Towers May 09 '21 at 22:31
  • @ahmad There are two methodologies I detailed in the referenced question (i e., contour integration, and the partial fraction expansion of the secant/cosecant function). In the answer I posted, I added a thorough treatise of use of the Beta and Gamma functions. – Mark Viola May 09 '21 at 22:56
  • This is just a Mellin transform of $\frac{1}{1+x}$. Ultimately most Mellin transform techniques I know are either residue based or Fourier/Laplace transform based (I'd argue a couple of those are Laplace transforms). So if you don't like the examples presented in the other posts, I'm not sure what to tell you. – Cameron Williams May 10 '21 at 13:57

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I'm not sure if this helps since in the end, I've reduced the problem to evaluating an infinite series (in which standard residue theorems are still required to evaluate the series).

This is done via a really simple decomposition of the integral:

\begin{equation} \begin{aligned} &\int^{\infty}_0 \frac{x^{p-1}}{1+x} dx \\ = &\int^{1}_0 \frac{x^{p-1}}{1+x} dx + \int^{\infty}_1 \frac{x^{p-1}}{1+x} dx \\ \end{aligned} \end{equation}

and we consider the infinite series

\begin{equation} \frac{1}{1+x} = \sum_{n=0}^\infty (-1)^n x^n \end{equation}

for $|x| < 1$ and

\begin{equation} \frac{1}{1+x} = \frac{1}{x}\frac{1}{1+1/x} = \sum_{n=0}^\infty (-1)^n x^{-n-1} \end{equation}

for $|x| > 1$.

The first integral can be evaluated as:

\begin{equation} \begin{aligned} &\int^{1}_0 \frac{x^{p-1}}{1+x} dx \\ = & \int^{1}_0 x^{p-1} \sum_{n=0}^\infty (-1)^n x^n dx \\ = & \sum_{n=0}^\infty (-1)^n \int^{1}_0 x^{p-1+n} dx \\ = & \sum_{n=0}^\infty \frac{(-1)^n}{n+p}. \\ \end{aligned} \end{equation}

Here, you can easily justify the interchange between the integral and infinite series using Fubini's Theorem.

Similarly, we have:

\begin{equation} \begin{aligned} &\int^{\infty}_1 \frac{x^{p-1}}{1+x} dx \\ = & \sum_{n=0}^\infty \frac{(-1)^n}{n+1-p}. \\ \end{aligned} \end{equation}

The two series can be combined to give:

\begin{equation} \begin{aligned} &\int^{\infty}_0 \frac{x^{p-1}}{1+x} dx \\ = &\frac{1}{p} + \sum_{n=1}^\infty \frac{(-1)^n}{n+p} - \sum_{n=1}^\infty \frac{(-1)^n}{n-p}\\ = &\frac{1}{p} - (2p) \sum_{n=1}^\infty \frac{(-1)^n}{n^2-p^2}.\\ \end{aligned} \end{equation}

At this point, you can look up a table of infinite series or use Mathematica to see that

\begin{equation} \sum_{n=1}^\infty \frac{(-1)^n}{n^2-p^2} = -\frac{\pi}{2p}\left(\frac{1}{p\pi} + \frac{1}{\sin(p\pi)} \right). \end{equation}

Substitute the above equation right in to obtain the required expression.

Remark: The way to evaluate the infinite series is usually done via Residue Theorem, so you can't really escape from that, although you could argue that you can search up a table of infinite series to obtain the results that you want.

HK Tan
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