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I looked around and found that integrals of the form

$$\int_{0}^{\infty} \frac{x^{m-1}}{a+x^n}, a,m,n \in \mathbb{R}, 0<m<n, 0<a$$

seem to occur very often:

Just to give a few examples (the formula given below would solve them all right away):

How can I solve the integral $\int_{0}^{\infty} \frac{dt}{1+t^4}$?

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

Is this definite integral really independent of a parameter? How can it be shown?

Integrating $\int_0^\infty \frac{1}{x^2 + 2x + 2}$d$x$

Even more surprising was that there seems to be a (quite beautiful) closed form, namely:

$$\int_{0}^{\infty} \frac{x^{m-1}}{a+x^n}=\frac{\pi}{n} \left(\frac{1}{a}\right)^{1-\frac{m}{n}} \csc \left(m \cdot \frac{\pi }{n}\right)$$

(This result is from Mathematica). I tried to derive this result by integrating (brute force) but you get hypergeometric functions which I don't like. Therefore I would like to know if there is a straight-forward way to get this by hand.

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I'm not familiar with these techniques myself, but I remember seeing the case m = a = 1 done with both contour integration and gamma/beta functions plus the reflection formula. This can probably be done similarly? –  Lyrebird Jun 12 '11 at 9:09
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minor remark: you might want to demand $a > 0$ as well. –  Gerben Jun 12 '11 at 11:00
    
Yes I missed that –  Listing Jun 12 '11 at 11:17
    
@Listing Another way is putting $x^{m-1} = e^t$ to get an integral solvable by infinite series of $\dfrac{1}{1+e^t}$. It produces ($u = \dfrac{m}{n}$) $$\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{u + k}} = \frac{\pi }{{\sin \pi u}}} $$ which I still need to show is true. –  Pedro Tamaroff Feb 17 '12 at 22:26
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1 Answer

up vote 16 down vote accepted

Well there is actually a formula which I found in Gamelin's complex analysis book. This is problem number $7$ Exercise VII.4 Page No. 208. This works for the case $a=1$. So if you have $a=1$, then $$\int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ \text{dx} = \frac{\pi}{b \sin(\pi{a}/b)}, \quad 0 < a < b$$

Set $$I = \int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ \text{dx}$$ and integrate $$f(z) = \frac{z^{a-1}}{1+z^{b}} = \frac{|z|^{a-1} \cdot e^{i(a-1)\text{arg}(z)}}{1+|z|^{b}e^{ib\text{arg}(z)}}$$

Simple pole at $z_{1} = e^{\pi{i}/b}$ and hence $$\text{Res} \Biggl[\frac{z^{a-1}}{1+z^{b}}, e^{\pi{i}/b}\Biggr] = \frac{z^{a-1}}{bz^{b-1}}\Biggl|_{z =e^{\pi i / b}} = -\frac{1}{b}e^{\pi i a/b}$$

Integrate along $\gamma_{1}$, and let $R \to \infty$ and let $ \epsilon \to 0^{+}$. This gives, \begin{align*} \int\limits_{\gamma_{1}} f(z) \ dz & = \int\limits_{\gamma_{1}} \frac{|z|^{a-1} \cdot e^{i(a-1)\text{arg}(z)}}{1+|z|^{b}e^{ib\text{arg}(z)}} \ dz \\ &= \int\limits_{\epsilon}^{R} \frac{x^{a-1}}{1+x^{b}} \to \int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ dx =I \end{align*}

Integrate along $\gamma_{2}$, and let $R \to \infty$. This gives $0 < a < b$ and $$\Biggl|\int\limits_{\gamma_{2}} f(z) dz \Biggr| \leq \frac{R^{a-1}}{R^{b}-1} \cdot \frac{2\pi R}{b} \sim \frac{2 \pi}{b R^{b-a}} \to 0$$

Integrate along $\gamma_{3}$ and let $R \to \infty$ and $\epsilon \to 0^{+}$. This gives \begin{align*} \int\limits_{\gamma_{3}} f(z) \ dz &= \int\limits_{\gamma_{3}} \frac{|z|^{a-1} \cdot e^{i(a-1)\text{arg}(z)}}{1+|z|^{b}e^{ib\text{arg}(z)}} = \Biggl[\begin{array}{c} z=x e^{2\pi i/b} \\ dz=e^{2\pi i/b} \ dx \end{array}\Biggr] \\ &= \int\limits_{R}^{\epsilon} \frac{x^{a-1}e^{2\pi i(a-1)/b}}{1+x^{b}} \cdot e^{2\pi i b} \ dx \to \int\limits_{\infty}^{0} \frac{x^{a-1}e^{2\pi i(a-1)/b}}{1+x^{b}} \cdot e^{2\pi i b} \ dx \\ &= -e^{2\pi ia/b}I \end{align*}

Integrate along $\gamma_{4}$ and let $\epsilon \to 0^{+}$. This gives $0 < a <b$, $$\Biggl|\int\limits_{\gamma_{4}} f(z) \ dz \Biggr| \leq \frac{\epsilon^{a-1}}{1-\epsilon^{b}} \cdot \frac{2\pi\epsilon}{b} \sim \frac{2\pi\epsilon}{b} \to 0$$

Using the $\text{Residue Theorem}$ and letting $R \to \infty$ and $\epsilon \to 0^{+}$, we obtain that $$ I + 0 - e^{2\pi a/b}I + 0 = 2\pi i \cdot \Bigl(-\frac{1}{b} e^{\pi ia/b}\Bigr)$$ This yields, $$(e^{-\pi i a/b} - e^{\pi i a./b})I= -\frac{2\pi i}{b}$$ and hence solving for $I$, we have $$I= \frac{2\pi i}{b \cdot (e^{\pi ia/b} - e^{-\pi i a/b})}=\frac{\pi}{b \sin(\pi a/b)}$$

Actually I have taken this answer from, one of my posts: You can see that as well.

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why was this downvoted? –  Thomas Rot Jun 12 '11 at 9:10
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@Thomas: Perhaps, because, this doesn't exactly answer the question and only answers a small case. –  user9413 Jun 12 '11 at 9:12
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I feel this answer should remain because: 1) Methods involved in evaluating residue. 2) Some other people might use this and answer the OP's question. –  user9413 Jun 12 '11 at 9:13
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Argument of z and residue z respectively, I believe. –  Lyrebird Jun 12 '11 at 9:20
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For $a > 0$, let $y = x/a^{1/n}$ and so $x = a^{1/n} y$. We have: $$x^n = a y^n$$ $$x^{m-1} = a^{(m - 1)/n} y^{m - 1}$$ $$dx = a^{1/n} dy$$ And for the integral $$\int_0^\infty \frac {x^{m - 1}} {a - x^n} dx = a^{m/n - 1}\int_0^\infty \frac {y^{m - 1}} {1 - y^n} dy$$ –  AlbertH Jun 12 '11 at 9:36
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