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$$\int_0^\infty \frac{x^2}{x^4+1} \; dx $$

All I know this integral must be solved with beta function, but how do I come to the form $$\beta (x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\;dt \text{ ?}$$

hardmath
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Serj
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    Maybe you have to solve it with beta function, but that's your homework, and your problem. Most people here know better methods. Residues are most elegant, partial fraction decomposition works, and the latter is good exercise. Go ahead! –  Jul 02 '17 at 19:15
  • please check this question (https://math.stackexchange.com/questions/110457/closed-form-for-int-0-infty-fracxn1-xmdx/176216#176216) – W.R.P.S Jul 02 '17 at 19:15
  • i got it , thank you for your time :) . – Serj Jul 02 '17 at 19:23
  • Gocan: Since you originally apologized for the post, in your post (which does not belong in a question or answer), I take it you realize you've posted a "problem-statement-question", with no contribution from you, and recognized that a question such as yours (no context, just a "do it for me" problem), is lacking in what the site expects from askers. (Please explain, e.g., what you know about the beta function, and why you note it must be solved through it's use, as assigned to you?). Please edit your post to improve it, adding context along the lines I suggest here. – amWhy Jul 02 '17 at 20:42

4 Answers4

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By definition $$B(p, q) = \int_0^1x^{p-1}(1-x)^{q-1}\,dx.$$ By using shift $x = \frac{t}{1 + t}$ previous integral becomes $$B(p,q) = \int_0^{\infty}\frac{t^{p-1}}{(1 + t)^{p + q}}\,dt.$$

Thus we get that \begin{align} I & = \int_0^\infty \frac{x^2}{1 + x^4} \, dx = \{x^4 = t, dx = t^{\frac14 - 1} \, dt, x^2 = t^{\frac12}\} = \frac14\int_0^\infty \frac{t^{\frac12}t^{\frac14 - 1}}{1+t} \, dt \\[10pt] & = \frac14\int_0^\infty \frac{t^{\frac34 - 1}}{(1 + t)^{\frac34 + \frac14}} \, dt = \frac14 B\left(\frac14, 1 - \frac14\right) = \frac14 \frac{\Gamma(\frac14)\Gamma(\frac14 - 1)}{\Gamma(\frac14 + 1 - \frac14)} = \frac14 \frac{\Gamma(\frac14)\Gamma(\frac14 - 1)}1 \\[10pt] & = \frac14\frac{\pi}{\sin(\frac14\pi)} = \frac{\pi}{2\sqrt2}. \end{align} I hope I have not made any mistakes

Michael Hardy
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Nemanja Beric
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Hint: If you have to use Beta function, let $x^2 = \tan \theta$ and use the definition of Beta function in terms of sines and cosines.

amWhy
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jgsmath
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  • Could you state that definition here? – Michael Hardy Jul 02 '17 at 20:19
  • @MichaelHardy Somebody linked my reply to Wiki article on Beta function already. Thanks!! – jgsmath Jul 02 '17 at 21:41
  • Still, there is some interest in making this self-contained. I'll put it here: $$ \begin{align} \operatorname{B}(x,y) &= 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1} , d\theta, && \operatorname{Re}(x)>0,\ \operatorname{Re}(y)>0 \end{align} $$ – Michael Hardy Jul 02 '17 at 22:10
  • @MichaelHardy OK.. I will try to make my future questions/answers self contained as much as possible. I'm relatively new here, so thanks for the tips!! – jgsmath Jul 02 '17 at 22:15
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note that $$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$$

Dr. Sonnhard Graubner
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$$\begin{eqnarray*}I=\int_0^{+\infty}\frac{x^2\,dx}{x^4+1}\stackrel{\text{parity}}{=} \frac{1}{2} \int_{-\infty}^{+\infty}\frac{dx}{x^2+\frac{1}{x^2}}&=&\frac{1}{2} \int_{-\infty}^{+\infty}\frac{dx}{\left(x-\frac{1}{x}\right)^2+2} \\[8pt] (\text{by Glasser's Master Theorem})&=&\int_0^{+\infty}\frac{dx}{x^2+2}=\color{red}{\frac{\pi}{2\sqrt{2}}}.\end{eqnarray*} $$

As a reference, http://mathworld.wolfram.com/GlassersMasterTheorem.html.
Through the Beta function, by setting $\frac{1}{x^4+1}=u$, $$ I=\frac{1}{4}\int_{0}^{1}u^{-3/4}(1-u)^{-1/4}\,du=\frac{\Gamma\left(\frac{1}{4}\right)\,\Gamma\left(\frac{3}{4}\right)}{4}=\frac{\pi}{4\sin\frac{\pi}{4}} =\color{red}{\frac{\pi}{2\sqrt{2}}}$$ by the reflection formula for the $\Gamma$ function.

Jack D'Aurizio
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  • Explain the downvote since I do not see anything wrong in this approach. – Jack D'Aurizio Jul 02 '17 at 20:09
  • The question asks for a way to use the Beta function. – orion Jul 02 '17 at 20:09
  • @JackD'Aurizio : Is there some standard name for the substitution $u = x - \dfrac 1 x,$ and is there some account of its use in evaluating integrals in Wikipedia or MathWorld or some textbooks or scholarly articles? – Michael Hardy Jul 02 '17 at 20:18
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    @MichaelHardy: it is sometimes known as the Cauchy-Schlomilch substitution. – Jack D'Aurizio Jul 02 '17 at 20:20
  • @MichaelHardy: as a reference, https://arxiv.org/abs/1004.2445 – Jack D'Aurizio Jul 02 '17 at 20:23
  • Another downvote ?! This time, for what? – Jack D'Aurizio Jul 02 '17 at 20:35
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    I for one up-voted this answer. I also created a new Wikipedia article titled Glasser's master theorem. It could use the following improvements: $$ \begin{align} \bullet & \text{ some concrete examples;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \bullet & \text{ other articles linking to it;} \ \bullet & \text{ probably other things.} \end{align} $$ – Michael Hardy Jul 02 '17 at 21:05