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

Question

I found the method of partial fractions very laborious to solve this definite integral : $$\int_0^\infty \frac{\sqrt[3]{x}}{1 + x^2}\,dx$$

Is there a simpler way to do this ?

Answer 1

Perhaps this is simpler.

Make the substitution $\displaystyle x^{2/3} = t$. Giving us

$\displaystyle \frac{2 x^{1/3}}{3 x^{2/3}} dx = dt$, i.e $\displaystyle x^{1/3} dx = \frac{3}{2} t dt$

This gives us that the integral is

$$I = \frac{3}{2} \int_{0}^{\infty} \frac{t}{1 + t^3} \ \text{d}t$$

Now make the substitution $t = \frac{1}{z}$ to get

$$I = \frac{3}{2} \int_{0}^{\infty} \frac{1}{1 + t^3} \ \text{d}t$$

Add them up, cancel the $\displaystyle 1+t$, write the denominator ($\displaystyle t^2 - t + 1$) as $\displaystyle (t+a)^2 + b^2$ and get the answer.

Answer 2

Here is a different way I am quite fond of:

Call our integral I, that is set $$I=\int_0^\infty \frac{\sqrt[3]{x}}{1+x^2}\,dx$$ Let $u=1+x^{2}$ so that $du=2x \, dx$. Since $$\sqrt[3]{x} \, dx=\frac{1}{2}\frac{2x \, dx}{\sqrt[3]{x^{2}}}=\frac{1}{2}\frac{u}{\left({u-1}\right)^{\frac{1}{3}}}\,du$$ we have that$$I=\frac{1}{2}\int_{1}^{\infty}\frac{1}{u\left(u-1\right)^{\frac{1}{3}}}\,du.$$ Let $u=\frac{1}{v}$ so that this becomes $$\frac{1}{2}\int_{0}^{1}\frac{1}{\frac{1}{v}\left(\frac{1}{v}-1\right)^{\frac{1}{3}}}\frac{1}{v^{2}}\,dv=\frac{1}{2}\int_{0}^{1}v^{-\frac{2}{3}}\left(1-v\right)^{-\frac{1}{3}}\,dv=\frac{1}{2}\text{B}\left(\frac{1}{3},\frac{2}{3}\right)$$ where $\text{B}(x,y)$ is the beta function. Since $$\text{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ we have that $$I=\frac{1}{2}\frac{\Gamma(\frac{1}{3})\Gamma(\frac{2}{3})}{\Gamma(1)}.$$ Since $\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin\pi s}$ it follows that $$I=\frac{\pi}{2\sin\pi/3}=\frac{\pi}{\sqrt{3}}.$$ Hope that helps,

Also it is worth mentioning that numerically, Wolfram Alpha agrees with this answer.

Answer 3

By using techniques of complex analysis ($\text{Residue Theory}$) one can actually show that $$\int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ \text{dx} = \frac{\pi}{b \sin(\pi{a}/b)}, \qquad 0 < a <b$$

You can obtain the value of your $\text{Integral}$ by putting $a=\frac{4}{3}$ and $b=2$.

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 $$I= \frac{2\pi i}{b \cdot (e^{\pi ia/b} - e^{-\pi i a/b})}=\frac{\pi}{b \sin(\pi a/b)}$$

Answer 4

Using the same technique as in my previous answer we can generalize, and find the Mellin Transform:

Consider $$I(\alpha,\beta)=\int_{0}^{\infty}\frac{u^{\alpha-1}}{1+u^{\beta}}du=\mathcal{M}\left(\frac{1}{1+u^{\beta}}\right)(\alpha)$$ Let $x=1+u^{\beta}$ so that $u=(x-1)^{\frac{1}{\beta}}$. Then we have $$I(\alpha,\beta)=\frac{1}{\beta}\int_{1}^{\infty}\frac{(x-1)^{\frac{\alpha-1}{\beta}}}{x}(x-1)^{\frac{1}{\beta}-1}dx.$$ Setting $x=\frac{1}{v}$ we obtain $$I(\alpha,\beta)=\frac{1}{\beta}\int_{0}^{1}v^{-\frac{\alpha}{\beta}}(1-v)^{\frac{\alpha}{\beta}-1}dv=\frac{1}{\beta}\text{B}\left(-\frac{\alpha}{\beta}+1,\ \frac{\alpha}{\beta}\right).$$

Using the properties of the Beta and Gamma functions, this equals $$\frac{1}{\beta}\frac{\Gamma\left(1-\frac{\alpha}{\beta}\right)\Gamma\left(\frac{\alpha}{\beta}\right)}{\Gamma(1)}=\frac{\pi}{\beta\sin\left(\frac{\pi\alpha}{\beta}\right)}.$$

Answer 5

Let's generalize the problem. We will evaluate $$ \int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx. $$ Let $$y=\dfrac{1}{1+x^b}\quad\Rightarrow\quad x=\left(\dfrac{1-y}{y}\right)^{\large\frac1b}\quad\Rightarrow\quad dx=-\left(\dfrac{1-y}{y}\right)^{\large\frac1b-1}\ \dfrac{dy}{by^2}\ ,$$ then \begin{align} \int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx&=\int_0^1 y\left(\dfrac{1-y}{y}\right)^{\large\frac{a-1}b}\left(\dfrac{1-y}{y}\right)^{\large\frac1b-1}\ \dfrac{dy}{by^2}\\&=\frac1b\int_0^1y^{\large1-\frac{a}{b}-1}(1-y)^{\large\frac{a}{b}-1}\ dy, \end{align} where the last integral in RHS is Beta function. $$ \text{B}(x,y)=\int_0^1t^{\ \large x-1}\ (1-t)^{\ \large y-1}\ dt=\frac{\Gamma(x)\cdot\Gamma(y)}{\Gamma(x+y)}. $$ Hence \begin{align} \int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx&=\frac1b\int_0^1y^{\large1-\frac{a}{b}-1}(1-y)^{\large\frac{a}{b}-1}\ dy\\&=\frac1b\cdot\Gamma\left(1-\frac{a}{b}\right)\cdot\Gamma\left(\frac{a}{b}\right)\\&=\large{\color{blue}{\frac{\pi}{b\sin\left(\frac{a\pi}{b}\right)}}}. \end{align} The last part uses Euler's reflection formula for Gamma function provided $0<a<b$. Thus $$ \large\int_0^\infty\dfrac{\sqrt[3]{x}}{1+x^2}\ dx=\int_0^\infty\dfrac{x^{\large\frac43-1}}{1+x^2}\ dx=\frac{\pi}{2\sin\left(\frac{2\pi}{3}\right)}=\color{blue}{\frac\pi3\sqrt{3}}. $$

Answer 6

We can convert the integral into a more familiar $\sin(x)/x$ type form using $$ I=\int_0^\infty f(x) g(x)\; dx = \int_0^\infty \mathcal{L}[f(x)](s)\mathcal{L}^{-1}[g(x)](s)\;ds $$ $$ \mathcal{L}[x^{1/3}](s)=\frac{\Gamma(4/3)}{s^{4/3}} $$ $$ \mathcal{L}^{-1}\left[\frac{1}{1+x^2}\right](s) =\sin(s) $$ So $$ \int_0^\infty \frac{x^{1/3}}{1+x^2}\;dx =\Gamma\left(\frac{4}{3}\right)\int_0^\infty \frac{\sin(x)}{x^{4/3}}\;dx = \frac{3}{2}\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{4}{3}\right) = \frac{\pi}{\sqrt{3}} $$

Answer 7

We can use the following results $$\sum_{n=-\infty}^\infty(-1)^n\frac{1}{bn+a}=\frac{\pi}{b\sin\frac{a\pi}{b}}, \frac{1}{1-x}=\sum_{n=0}^\infty x^n $$ to evaluate the generalization. In fact \begin{eqnarray} \int_0^\infty\dfrac{x^{a-1}}{1+x^b}\ dx&=&\int_0^1\frac{x^{a-1}}{1+x^b}dx+\int_0^1\frac{x^{-a-1}}{1+x^b}dx\\ &=&\sum_{n=0}^\infty(-1)^n\int_0^1(x^{bn+a-1}+x^{bn-a-1})dx\\ &=&\sum_{n=0}^\infty(-1)^n(\frac{1}{bn+a}+\frac{1}{bn-a})\\ &=&\sum_{n=-\infty}^\infty(-1)^n\frac{1}{bn+a}\\ &=&\frac{\pi}{b\sin\frac{a\pi}{b}}. \end{eqnarray}

Answer 8

Another way

Let

$$ I= \int_0^\infty \frac {x^{1/3}\, \mathrm{d}x}{x^2 + 1}. $$

By the Schwinger parametrization we have

$$ I=\int_0^\infty \mathrm{d}t\, \exp(-t)\int_0^\infty \mathrm{d}x\, x^{1/3}\, \exp\left(-tx^2\right).$$

The last integral can be calculated along the lines of this answer. Using this result, one gets

$$I=\int_0^\infty \mathrm{d}t\,\frac{ \Gamma \left(\frac{2}{3}\right)\exp(-t)}{2 t^{2/3}}=\frac{1}{2}\Gamma \left(\frac{2}{3}\right)\Gamma \left(\frac{1}{3}\right)=\frac{\pi}{\sqrt{3}},$$

since

$$ \Gamma\left(\nu,x\right)\equiv\int_x^\infty \mathrm{d}t\, t^{\nu-1}\exp(-t).$$