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I have a question which asks to find the value of:

$\displaystyle \tag*{} \int \limits _{0}^{\infty} \dfrac{\mathrm dx}{\sqrt{x^n+a} + \sqrt{x^n+b}}$

Where, $a,b >0$ and $n >2$

I tried to rationalize the denominator and arrived:

$\displaystyle \tag*{} \dfrac{1}{a-b}\int \limits _{0}^{\infty}\sqrt{x^n+a} - \sqrt{x^n+b} \ \ \mathrm dx $

Since individual integrals do not converge. But couldn't proceed any further. Any help would be greatly appreciated. Thank you!

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1 Answer 1

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We'll evaluate the following auxiliary integral:

$$\int_0^\infty \sqrt{x^n + \alpha} - x^{\frac{n}{2}} \, \mathrm{d}x = \frac{\alpha^{\frac1n + \frac12} }{(n+2)\sqrt{\pi}}\Gamma\left(\frac{1}{2} - \frac{1}{n} \right)\Gamma\left( \frac{1}{n} \right)\qquad \text{for} \quad \alpha >0, \, n>2$$

Proof: Taking the substitution $\sqrt{x^n +\alpha}- \sqrt{x^n} = \sqrt{\alpha t}$ gives $x = \alpha^{\frac1n}2^{-\frac{2}{n}}t^{-\frac{1}{n}}\left(1-t\right)^{\frac{2}{n}}$ which then gives $\mathrm{d}x =-\frac{1}{n}\alpha^{\frac1n}2^{-\frac{2}{n}}\left[ t^{-\frac1n -1}(1-t)^{\frac2n -1} + t^{-\frac1n}(1-t)^{\frac2n -1}\right] \mathrm{d}t $. Combining everything we get: $\require{cancel}$ \begin{align*} \int_0^\infty \sqrt{x^n + \alpha} - x^{\frac{n}{2}} \, \mathrm{d}x & =\frac{\alpha^{\frac1n + \frac12}2^{-\frac{2}{n}}}{n}\left[\int_{0}^{1} t^{\frac12-\frac1n -1}(1-t)^{\frac2n -1}\mathrm{d}t + \int_{0}^{1} t^{\frac12-\frac1n +1-1}(1-t)^{\frac2n -1}\mathrm{d}t\right]\\ & =\frac{\alpha^{\frac1n + \frac12}2^{-\frac{2}{n}}}{n}\left[B\left(\frac12-\frac1n, \frac2n\right) + B\left(\frac12-\frac1n+1, \frac2n\right)\right]\\ & =\frac{\alpha^{\frac1n + \frac12}2^{-\frac{2}{n}}}{n}\left[1 + \frac{\frac12 - \frac1n}{\frac12+\frac1n}\right]B\left(\frac12-\frac1n, \frac2n\right)\\ & =\frac{\alpha^{\frac1n + \frac12}\cancel{2^{-\frac{2}{n}}}}{\cancel{n}}\frac{\cancel{2}\cancel{n}}{n+2}\frac{\cancel{2^{\frac2n-1}}\cancel{\Gamma\left(\frac1n + \frac12\right)}\cancel{\Gamma\left(\frac12\right)}}{\cancel{\sqrt{\pi}}\cancel{\Gamma\left(\frac12 + \frac1n\right)}}B\left(\frac12-\frac1n, \frac1n\right)\\ & =\frac{\alpha^{\frac1n + \frac12} }{n+2}\frac{\Gamma\left(\frac{1}{2} - \frac{1}{n} \right)\Gamma\left( \frac{1}{n} \right)}{\sqrt{\pi}} \end{align*} using that $B(x+1,y) = \frac{x}{x+y}B(x,y)$ and $B(x,2y) = \frac{2^{2y-1}\Gamma\left(y + \frac12\right)\Gamma\left(x+y\right)}{\sqrt{\pi} \,\Gamma\left(x+2y\right)}B(x,y)$. Note that the first result is a consequence of $\Gamma(z+1) = z\Gamma(z)$ and the second result is a consequence of Legendre's duplication formula.


With the previous result we can conclude the problem as follows: \begin{align*} I &=\frac{1}{a-b} \int_{0}^{\infty} \sqrt{x^n +a} -\sqrt{x^n +b} \, \mathrm{d}x\\ & = \frac{1}{a-b} \left[\int_{0}^{\infty} \sqrt{x^n +a}\color{blue}{-x^{\frac{n}{2}}} \, \mathrm{d}x-\int_{0}^{\infty}\sqrt{x^n +b}\color{blue}{-x^{\frac{n}{2}}} \, \mathrm{d}x\right]\\ & = \boxed{\frac{a^{\frac1n + \frac12 } - b^{\frac1n + \frac12 } }{(a-b)(n+2)\sqrt{\pi}} \Gamma \left(\frac12 - \frac1n\right) \Gamma \left( \frac1n\right)} \end{align*}

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    $\begingroup$ I think you were moving in the right direction, but the integral $\int_{0}^{1} \left(\frac{t}{\alpha}\right)^{-\frac12}t^{-\frac1n -1}(1-t)^{\frac1n -1}\mathrm{d}t$ evidently diverges at zeto. In my opinion, more justification of this step is needed. $\endgroup$
    – Svyatoslav
    Mar 10, 2022 at 5:03
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    $\begingroup$ @Svyatoslav, thank you for pointing that out! I've written a new derivation to circumvent the convergence issue. $\endgroup$
    – Robert Lee
    Mar 10, 2022 at 10:18
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    $\begingroup$ excellent intuition (+1) $\endgroup$
    – G Cab
    Mar 10, 2022 at 10:24

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