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How would you compute for the definite integral of $$\int_0^{\infty}\frac{dx}{(1+x^2)^4}$$ I know that integral of $\displaystyle \frac1{(1+x^2)}$ equals $\tan^{-1}x$. I tried using integration by parts without much luck. My teacher pointed me to special functions by which I found out about the hypergeometric distribution. Although I don't know how to apply it to this problem. Anybody know how to use special functions or how to go about this problem? |
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The method given below works in general for any integral of the form $\displaystyle \int_{0}^{\infty} \frac{dx}{(1+x^2)^n}$. Plug in $x = \tan(\theta)$. The integral becomes $$\displaystyle \int_{0}^{\infty} \frac{dx}{(1+x^2)^4} = \int_{0}^{\pi/2} \frac{\sec^2(\theta) d \theta}{(1+\tan^2(\theta))^4} = \int_{0}^{\pi/2} \cos^{6}(\theta) d\theta = \frac{5}{6}\frac{3}{4}\frac{1}{2}\frac{\pi}{2} = \frac{5}{32}\pi$$ where the last integral has been done in a previous post here. The integration over there has been done for $\sin^n(\theta)$ but the same method works for $\cos^n(\theta)$. If you are familiar with complex analysis, you could solve it using complex analysis by extending the function into the complex domain, choose a semicircular contour with the diameter along the real axis. Look for the poles inside the contour, (there are four poles at $z=+i$) and compute the residues. Let the radius of the semicircle tend to infinity and evaluate the integral using Cauchy residue theorem. |
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Integration by parts does in fact help. Let $I_n = \int (1+x^2)^{-n}dx$. Then multiply by 1 and integrate by parts: $$I_n = x (1+x^2)^{-n} - \int x \cdot (-n)2x (1+x^2)^{-n-1}dx = \dots = x (1+x^2)^{-n} + 2n(I_n - I_{n+1}).$$ (In the "..." step, write $x^2=(x^2+1)-1$ in the numerator and then divide.) From this you can solve for $I_{n+1}$ in terms of $I_n$, and since you know $I_1 = \arctan x + C$, you can recursively compute $I_2$, $I_3$, $I_4$, etc. (This is pretty much the same method that Sivaram pointed you to for finding $\int \cos^6 \theta \, d\theta$.) |
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