How to solve this integral using residue theorem?

$$\int_0^∞ \frac{x^{(a-1)}}{(x+b)(x+c)} \, dx $$

$0 < a < 1, \ \ \ b > 0, \ \ \ c > 0$


closed as off-topic by user223391, Omnomnomnom, Jack D'Aurizio, Hurkyl, saz Oct 2 '16 at 6:31

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  • 2
    $\begingroup$ What are your thoughts on the question? What have you tried? Did you come up with a contour that you think will work? Please note for this and future questions, askers are expected to provide some context with their questions. $\endgroup$ – Omnomnomnom Oct 1 '16 at 18:53
  • $\begingroup$ Can't you simply use the residues laws? For example, $a$ and $b$ are just numbers. They are also the poles of the denominators... which means... $\endgroup$ – Von Neumann Oct 1 '16 at 18:54
  • $\begingroup$ Also, note that this integral is not convergent in the usual sense. However, as is often the case in complex analysis, we're looking for the CPV. $\endgroup$ – Omnomnomnom Oct 1 '16 at 18:55
  • $\begingroup$ I can't think of any contour as poles will lie on the real axis. Can I take a semi-circle with origin as center and passing through Ri and -Ri? $\endgroup$ – Sherlock Oct 1 '16 at 18:57
  • $\begingroup$ @omnomnomnom The integral exists an an improper Riemann integral and as a Lebesgue integral. There is no need to interpret it as a Cauchy Principal value. Why did you believe otherwise. $\endgroup$ – Mark Viola Oct 2 '16 at 12:51

Herein, we present an approach that exploits the residue theorem. Let $I(a,b,c)$ be the integral given by

$$\bbox[5px,border:2px solid #C0A000]{I(a,b,c)=\int_0^\infty \frac{x^{a-1}}{(x+b)(x+c)}\,dx} \tag 1$$

where $0<a<1$, $b>0$, and $c>0$.

Next, we move to the complex plane and evaluate the closed contour integral $J(a,b,c)$ given by

$$\bbox[5px,border:2px solid #C0A000]{J(a,b,c)=\oint_C \frac{z^{a-1}}{(z+b)(z+c)}\,dz} \tag2$$

where we choose the branch cut that emanates from the branch point at $z=0$ and extends along the positive real axis to $z=\infty$.

In $(2)$, the contour $C$ is the classical "keyhole" contour comprised of $(i)$ the ray from $\epsilon>0$ to $R>\max(b,c)$ on the upper side of the branch cut, $(ii)$ the counter clockwise circular path on which $z=Re^{i\phi}$, from $\phi =0$ to $\phi =2\pi$, $(iii)$ the ray from $R$ to $\epsilon$ on the lower side of the branch cut, and $(iv)$ the clockwise circular path on which $z=\epsilon e^{i\phi}$, from $\phi=2\pi$ to $\phi=0$.

From the residue theorem, we have

$$\begin{align}J(a,b,c)&=2\pi i \text{Res}\left( \frac{z^{a-1}}{b+c},z=-b,-c\right)\\\\ &=2\pi i \left( \frac{(-b)^{a-1}-(-c)^{a-1}}{c-b}\right)\\\\ &=\bbox[5px,border:2px solid #C0A000]{2\pi i e^{i\pi a}\left(\frac{b^{a-1}-c^{a-1}}{b-c}\right)}\tag 3 \end{align}$$

Next, we express $J(a,b,c)$ as the sum of integrals over the four contours that comprise $C$. Proceeding, we write

$$\begin{align} J(a,b,c)&=\int_\epsilon^R \frac{x^{a-1}}{(x+b)(x+c)}\,dx+\int_0^{2\pi}\frac{iR^{a}e^{ia\phi}}{(Re^{i\phi}+b)(Re^{i\phi}+c)}\,d\phi\\\\ &+\int_R^\epsilon \frac{x^{a-1}e^{i2\pi(a-1)}}{(x+b)(x+c)}\,dx+\int_{2\pi}^0\frac{i\epsilon^{a}e^{ia\phi}}{(\epsilon e^{i\phi}+b)(\epsilon e^{i\phi}+c)}\,d\phi \tag 3 \end{align}$$

As $R\to \infty$ and $\epsilon\to0$, the second and fourth integrals on the right-hand side of $(3)$ approach $0$. Therefore, we can write

$$\bbox[5px,border:2px solid #C0A000]{\lim_{\epsilon \to 0,R\to \infty}J(a,b,c)=(1-e^{i2\pi (a-1)})\int_0^\infty \frac{x^{a-1}}{(x+b)(x+c)}\,dx }\tag 4$$

Putting together $(3)$ and $(4)$ yields

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{x^{a-1}}{(x+b)(x+c)}\,dx=-\frac{\pi}{\sin(\pi a)}\,\frac{b^{a-1}-c^{a-1}}{b-c}}$$

  • $\begingroup$ I don't agree, you (tried to) use the residue theorem on a Riemann surface (where the branch point disappears). What I did below uses the standard residue theorem. $\endgroup$ – reuns Oct 1 '16 at 19:47
  • $\begingroup$ @user1952009 The path is a keyhole contour. The branch point is at $z=0$ and the cut is along to non-negative real axis. The Residue Theorem was applied correctly and the result matches yours. Wha t is the objection? $\endgroup$ – Mark Viola Oct 1 '16 at 19:57
  • $\begingroup$ $z^{a-1}$ has a branch point at $z = 0$. It is analytic on a (curved) Riemann surface not on $\mathbb{C}$ $\endgroup$ – reuns Oct 1 '16 at 19:58
  • $\begingroup$ And as you noticed, the result doesn't match $\endgroup$ – reuns Oct 1 '16 at 20:00
  • $\begingroup$ So, at least one is wrong. $\endgroup$ – Mark Viola Oct 1 '16 at 20:01

The branch point at $x= 0$ tell us to use $x = e^u$ : $$I = \int_0^\infty \frac{x^{a-1}}{(x+b)(x+c)}dx = \int_{-\infty}^\infty \frac{e^{au}}{(e^u +b)(e^u+c)}du$$ Since $e^{2i \pi a}I =\int_{2i\pi-\infty}^{2i\pi+\infty} \frac{e^{au}}{(e^u +b)(e^u+c)}du$ you have with $C_R$ the rectangular contour $-R \to R\to R+2i\pi\to R-2i\pi \to-R$ and assuming $b > 0,c> 0$ : $$I\frac{1-e^{2i \pi a}}{2i\pi} = \lim_{R \to \infty} \frac{1}{2i\pi}\int_{C_R} \frac{e^{au}}{(e^u +b)(e^u+c)}du $$ $$ = Res( \frac{e^{au}}{(e^u +b)(e^u+c)},\log(b)+i\pi)+Res( \frac{e^{au}}{(e^u +b)(e^u+c)},\log(c)+i\pi)$$ $$ = \frac{e^{a(\log(b)+i\pi)}}{e^{\log(b)+i\pi}(e^{\log(b)+i\pi}+c)}+\frac{e^{a(\log(c)+i\pi)}}{e^{\log(c)+i\pi}(e^{\log(c)+i\pi}+b)} = e^{i \pi a}\frac{b^{a-1}-c^{a-1}}{b-c} $$

By analytic continuation it stays true for every $b,c \in \mathbb{C} \setminus (-\infty,0]$

  • $\begingroup$ how did you calculate poles to be b+iπ and c+iπ? e^{u}+b = e^{b+iπ}+b = -e^{b}+b which is not zero. $\endgroup$ – Sherlock Oct 1 '16 at 19:37
  • $\begingroup$ @Sherlock tell me if you find another mistake :) $\endgroup$ – reuns Oct 1 '16 at 19:43
  • $\begingroup$ In your final result, $a$ should be $a-1$ for the exponents, I believe. $\endgroup$ – Mark Viola Oct 1 '16 at 19:51
  • $\begingroup$ @Dr.MV ok find the mistake in my derivation $\endgroup$ – reuns Oct 1 '16 at 19:53
  • $\begingroup$ @Dr.MV So no, I think when you try to use the residue theorem on a Riemann surface, you forget that the curvature of the Riemann surface changes the residues at $x = b, x =c$ $\endgroup$ – reuns Oct 1 '16 at 19:55

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{\infty}{x^{a - 1} \over \pars{x + b}\pars{x + c}} \,\dd x:\ ?.\qquad 0 < a <1\,,\quad b >0\,,\quad c>0}$.

$$\bbox[8px,#efe,border:0.1em groove navy]{\mbox{It's interesting to see that the integral can be performed by 'real methods'}}$$

\begin{align} &\color{#f00}{\int_{0}^{\infty}{x^{a - 1} \over \pars{x + b}\pars{x + c}}\,\dd x} = {1 \over c - b}\pars{\int_{0}^{\infty}{x^{a - 1} \over x + b}\,\dd x - \int_{0}^{\infty}{x^{a - 1} \over x + c}\,\dd x}\label{1}\tag{1} \end{align}

  • The LHS integral converges whenever $\ds{0 < \Re\pars{a} < \color{#f00}{2}}$ albeit the OP asked for the condition $\ds{0 < a < \color{#f00}{1}}$.
  • The RHS integrals converge whenever $\ds{0 < \Re\pars{a} < \color{#f00}{1}}$ which coincides with the OP condition $\ds{0 < a < 1}$ whenever $\ds{a \in {\mathbb R}}$.
  • $\bbox[8px,#ffe,border:0.1em groove navy]{\mbox{Then,}\ \eqref{1}\ \mbox{is evaluated with}\ \ds{0 < \Re\pars{a} < \color{#f00}{1}}\ \mbox{which is more general that the above OP condition}}$

In the first integral we make $\ds{x/b \mapsto x}$ while in the second we make $\ds{x/c \mapsto x}$: \begin{align} &\color{#f00}{\int_{0}^{\infty}{x^{a - 1} \over \pars{x + b}\pars{x + c}}\,\dd x} = {b^{a - 1} - c^{a - 1} \over c - b}\int_{0}^{\infty}{x^{a - 1} \over x + 1} \,\dd x\label{2}\tag{2} \end{align}
Now, $\ds{t \equiv {1 \over x + 1}\implies x = {1 \over t} - 1\implies \totald{x}{t} = -\,{1 \over t^{2}}}$: \begin{align} &\color{#f00}{\int_{0}^{\infty}{x^{a - 1} \over \pars{x + b}\pars{x + c}}\,\dd x} = {b^{a - 1} - c^{a - 1} \over c - b}\int_{1}^{0}t\pars{{1 \over t} - 1}^{a - 1} \,\pars{-\,{\dd t \over t^{2}}} \\[5mm] = &\ {b^{a - 1} - c^{a - 1} \over c - b}\int_{0}^{1}t^{-a}\pars{1 - t}^{a - 1}\,\dd t \label{3.a}\tag{3.a} = {b^{a - 1} - c^{a - 1} \over c - b}\, {\Gamma\pars{-a + 1}\Gamma\pars{a} \over \Gamma\pars{1}} \\[5mm] = &\ \color{#f00}{{b^{a - 1} - c^{a - 1} \over c - b}\,{\pi \over \sin\pars{\pi a}}} \label{3.b}\tag{3.b} \end{align}

Note that the integral involved in \eqref{3.a} $\pars{~the\ Beta\ function~}$ converges whenever $$ \Re\pars{-a} > -1\,,\quad\Re\pars{a - 1} > -1\qquad\implies\qquad 0 < \Re\pars{a} < 1 $$ which defines the general condition for the integrals evaluation.

  • $\begingroup$ @Dr.MV I'm checking. Lets see$\dots$. Thanks. $\endgroup$ – Felix Marin Oct 2 '16 at 4:44
  • $\begingroup$ @Dr.MV Thanks. I add some comments to clarify the whole situation. $\endgroup$ – Felix Marin Oct 2 '16 at 5:07

You can use the result:

$$\int_0^\infty \frac{x^{-p}}{1+x}dx = \frac{\pi}{\sin{\pi p}}\tag{1}$$

To prove this result, you can define $z^{-p}$ as $r^{-p}\exp(-i\pi p\theta)$ where $z = r\exp(i\theta)$ is the polar representation of $z$, where we choose $0\leq\theta<2\pi$ so that the branch cut is put on the positive real axis. Then consider the contour that starts on the real axis at $z = \epsilon$ moves to $z =R$ on the real axis, then it moves via a counterclockwise circle of radius $R$ and center the origin to just below the real axis so that we just don't cross the branch cut, we then move infinitesimally below the positive real axis to just below $z = \epsilon$, and then we close the contour via a clockwise circle of radius $\epsilon$ with center the origin.

There is then one pole at $z = -1$ inside the contour, while the branch point singularity at $z = 0$ is outside the contour, the contour integral can thus be calculated using the residue method.

To calculate the integral in the question, you can expand


in partial fractions and then use (1) for each term (you don't have the re-calculate the contour integral when $x+1$ in the numerator is changed to $x+b$ as you can substitute $x = b t$ and then factor out powers of $b$).

Convergence issues are not a problem either when using (1) term by term for the partial fractions to get the desired result which has a different range of convergence for the parameter $a$. You can invoke the principle of analytic continuation to prove the result for $0<a<1$


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