Question

Prove that \(I_{n}=\left(\frac{n-1}{n+2}\right) I_{n-1},\) where \(I_{n}=\int_{0}^{\infty} \frac{x^{2 n-1}}{\left(x^{2}+1\right)^{n+3}} d x, \quad n \geq 1 .\) Then evaluate each integral. (a) \(\int_{0}^{\infty} \frac{x}{\left(x^{2}+1\right)^{4}} d x\) (b) \(\int_{0}^{\infty} \frac{x^{3}}{\left(x^{2}+1\right)^{5}} d x\) (c) \(\int_{0}^{\infty} \frac{x^{5}}{\left(x^{2}+1\right)^{6}} d x\)

Short answer

\nThe results of the integral calculations are as follow:\n(a) \(I_{1} = \frac{\pi}{12}\) \n(b) \(I_{2} = \frac{\pi}{60}\) \n(c) \(I_{3} = \frac{\pi}{280}\)

Step-by-step solution

Showing the Initial Relation

To show that \(I_{n}=\left(\frac{n-1}{n+2}\right) I_{n-1}\), it's essential to apply integration by parts. Use the formula \(\int udv = uv - \int vdu\), let \(u = x^{2n-1}\), \(dv = \frac{1}{{(x^{2}+1)}^{n+3}} dx\), hence, \(du = (2n-1)x^{2n-2} dx\) and \(v = -\frac{1}{2(n+2)}(x^{2}+1)^{n+2}\). With these substitutions perform integration by parts.

Integration By Parts

Substitute \(u\), \(v\), \(du\) and \(dv\) into the formula - \(\int udv = uv - \int vdu\). This gives \(I_{n} = u*v |_{0}^{\infty} - \int_{0}^{\infty} vdu \). Evaluating \(uv\) from \(0\) to \(\infty\) and simplifying the second integral should lead to the proposed equation.

Evaluating Integral (a)

For the integral \(\int_{0}^{\infty} \frac{x}{(x^{2}+1)^{4}} dx \), recognize it as \(I_{1}\). From the established formula you know that \(I_{1}=\frac{1}{3} I_{0}\). The tricky part here is to identify that \(I_{0} = \frac{\pi}{4}\) which is a standard integral. So plug in \(I_{0}\) to find \(I_{1}\).

Evaluating Integral (b)

For the integral \(\int_{0}^{\infty} \frac{x^{3}}{(x^{2}+1)^{5}} dx \), identify it as \(I_{2}\). Plug in \(n = 2\) to the relation to find \(I_{2}\) given that \(I_{1}\) was found in the previous step.

Evaluating Integral (c)

For the integral \(\int_{0}^{\infty} \frac{x^{5}}{(x^{2}+1)^{6}} dx \), identify it as \(I_{3}\). Plug in \(n = 3\) into the relationship to find \(I_{3}\) given that \(I_{2}\) has been found in the previous step.