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I am seeking assistance in evaluating the following definite integral:

$$\int_{-\infty}^{\infty} \frac{1}{(𝑥 ^2 + 1)^3} 𝑑x$$

Contour of the following integral

I have attempted various approaches, but I haven't been successful in finding a closed-form solution.

Could someone provide guidance on how to approach this integral and possibly compute its value? Any insights, alternative methods, or relevant references would be greatly appreciated.

Thank you in advance for your assistance!

Feel free to customize the question based on any specific details or context you'd like to add.

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3 Answers 3

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\begin{align} &\int_{-\infty}^{\infty} \frac{1}{(𝑥 ^2 + a)^3} 𝑑x \\ =&\ \frac12\frac{d^2}{da^2} \left(\int_{-\infty}^{\infty} \frac{1}{𝑥 ^2 + a} 𝑑x\right)_{a=1} = \frac12\frac{d^2}{da^2} \left(\frac\pi{\sqrt a}\right)_{a=1}= \frac{3\pi}8 \end{align}

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  • $\begingroup$ Thanks for the answer $\endgroup$ Commented Dec 13, 2023 at 15:55
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Since given function is even, we can just evaluate the integral from 0 to infinity and double it.

Let $\ t = \frac{1}{1+x^2}$.

Then, $\ dx = \frac{-dt}{2t^{\frac{3}{2}}(1-t)^{\frac{1}{2}}}$

Substituting $\ t$ throughout, we get:

$$\ I = \int_{\ 1}^{0} t^3 \cdot \frac{-dt}{t^{\frac{3}{2}}(1-t)^{\frac{1}{2}}}$$

$$ = \int_{\ 0}^{\ 1} t^{\frac{3}{2}}(1-t)^{\frac{-1}{2}}dt$$ $$ = \ B(\frac{5}{2},\frac{1}{2})$$

Where $\ B(x,y)$ is the Beta function.

Evaluating, we get

$$\ I = \frac{\Gamma(\frac{5}{2})\Gamma(\frac{1}{2})}{\Gamma(3)}$$

$$\ I = \frac{\frac{3\sqrt{\pi}}{4} \cdot \sqrt{\pi}}{2}$$

$$\ I = \frac{3\pi}{8}$$

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HINT…try substituting $$x=\tan\theta.$$ The resulting integral is fairly straightforward.

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    $\begingroup$ Especially if you know$$8\cos^4\theta=3+4\cos2\theta+\cos4\theta.$$ $\endgroup$
    – J.G.
    Commented Dec 12, 2023 at 13:00
  • $\begingroup$ Thanks for the hint!!! $\endgroup$ Commented Dec 13, 2023 at 15:55

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