# $I_n(t,a) = \int_0^\infty \frac{\cos(xt)}{\left(x^2 + a^2\right)^n}\:dx$

Spurred on by this, here I'm hoping to resolve the following integral: $$$$I_n(a,t) = \int_0^\infty \frac{\cos(xt)}{\left(x^2 + a^2\right)^n}\:dx$$$$ Where $$a,t \in \mathbb{R}^+$$ and $$n \in \mathbb{N}$$. To begin with we observe that: $$$$I_n(a,t) = \int_0^\infty \frac{\cos(xt)}{\left(a^2\left(\frac{x^2}{a^2} + 1\right)\right)^n}\:dx = \frac{1}{a^{2n}} \int_0^\infty \frac{\cos(xt)}{\left(\left(\frac{x}{a}\right)^2 + 1\right)^n}\:dx$$$$ Let $$u = \frac{x}{a}$$: \begin{align} I_n(a,t) &= \frac{1}{a^{2n}} \int_0^\infty \frac{\cos(uat)}{\left(u^2 + 1\right)^n}\cdot a\:du = a^{1 - 2n}\int_0^\infty \frac{\cos(uat)}{\left(u^2 + 1\right)^n}\:du \\ &=a^{1 - 2n}I_n(1, at) \end{align} Thus, we need only resolve the following integral to solve $$I_n(a,t)$$: $$$$J_n(s) = \int_0^\infty \frac{\cos(su)}{\left(u^2 + 1\right)^n}\:du$$$$ Noting $$I_n(a,t) = J_n(at)$$. Here we will proceed by forming a differential equation for $$J_n(s)$$. To do so, we employ Leibniz's Integral Rule and differentiate under the curve twice w.r.t $$s$$: \begin{align} \frac{d^2J_n}{ds^2} &= \int_0^\infty \frac{-u^2\cos(su)}{\left(u^2 + 1\right)^n}\:du = -\int_0^\infty \frac{\left(u^2 + 1 - 1\right)\cos(su)}{\left(u^2 + 1\right)^n}\:du \nonumber \\ &=-\left[\int_0^\infty \frac{\cos(su)}{\left(u^2 + 1\right)^{n - 1}}\:du - \int_0^\infty \frac{\cos(su)}{\left(u^2 + 1\right)^n}\:du\right] \nonumber \\ &=-\left[J_{n - 1}(s) - J_n(s) \right] = J_n(s) - J_{n - 1}(s) \end{align} Thus we form the recursive differential equation: $$$$\frac{d^2J_n}{ds^2}- J_n(s) = -J_{n - 1}(s)$$$$ In order for a solution to be obtained, the following is required: $$I_1(s)$$, $$I_n(0)$$, and $$I_n'(0)$$. Thankfully these are all easy to obtain. Starting with $$I_1(s)$$ we find: $$$$I_n(s) = \frac{\pi}{2}e^{-s}$$$$ For $$I_n(0)$$ we have: $$$$I_n(0) = \int_0^\infty \frac{1}{\left(u^2 + 1\right)^n}\:du$$$$ Using the subsitution $$u = \tan(w)$$ we obtain a solution in terms of the Beta (and by extension Gamma) function: \begin{align} I_n(0) &= \int_0^\frac{\pi}{2} \frac{1}{\left(\tan^2(w) + 1\right)^n}\cdot \sec^2(w)\:dw = \int_0^\frac{\pi}{2} \cos^{2n - 2}(w)\:dw \nonumber \\ &= \frac{1}{2}B\left( \frac{2n - 1}{2}, \frac{1}{2} \right) = \frac{1}{2}\frac{\Gamma\left(\frac{2n - 1}{2}\right)\Gamma\left( \frac{1}{2} \right)}{\Gamma\left(\frac{2n - 1}{2} + \frac{1}{2} \right)} = \frac{\sqrt{\pi}}{2}\frac{\Gamma\left(\frac{2n - 1}{2}\right)}{\Gamma(n)} \end{align} For $$I_n'(0)$$ we have: $$$$I_n'(0) = \int_0^\infty \frac{-x\sin(x \cdot 0)}{\left(x^2 + 1\right)^n} = 0$$$$ Now, and here is where I'm unsure about my process - for our recursive differential equation we take the Laplace Transform: \begin{align} \mathscr{L}_{s \rightarrow p}\left[ \frac{d^2J_n}{ds^2} \right] - \mathscr{L}_{s \rightarrow p}\left[J_n(s) \right] &= -\mathscr{L}_{s \rightarrow p}\left[ J_{n - 1}(s) \right] \nonumber \\ p^2 \overline{J}_n(p) - pJ_n(0) - J_n'(0) - \overline{J}_{n}(p) &= -\overline{J}_{n - 1}(p) \nonumber \\ \left(p^2 - 1\right)\overline{J}_n(p) &= pJ_n(0) -\overline{J}_{n - 1}(p) \end{align} Thus, $$$$\overline{J}_n(p) = \frac{p}{p^2 - 1} J_n(0) - \frac{1}{p^2 - 1}\overline{J}_{n - 1}(p)$$$$ We now take the Inverse Laplace Transform: \begin{align} \mathscr{L}_{p \rightarrow s}^{-1} \left[\overline{J}_n(p)\right] &= \mathscr{L}_{p \rightarrow s}^{-1} \left[\frac{p}{p^2 - 1}\right]J_n(0) - \mathscr{L}_{p \rightarrow s}^{-1} \left[\frac{1}{p^2 - 1}\overline{J}_{n - 1}(p)\right] \nonumber \\ J_n(s) &= J_n(0)\cosh(s) - \int_0^s \sinh(s - a)J_{n - 1}(a)\:da \nonumber \\ &= J_n(0)\cosh(s) - \int_0^s \left[\sinh(s)\cosh(a) - \sinh(a)\cosh(s)\right]J_{n - 1}(a)\:da \nonumber \\ &= J_n(0)\cosh(s) - \sinh(s)\int_0^s\cosh(a) J_{n - 1}(a)\:da \nonumber \\ &\quad+ \cosh(s)\int_0^2 \sinh(a)J_{n - 1}(a)\:da \end{align} Now whilst we have a recursive integral form that governs $$J_n(s)$$ I am unsure how to solve it!.

Does anyone have any pointers about how to move forward?

Another approach (I believe) is to employ the linear D-operator. Here if we define $$D = \frac{d}{ds}$$ then our governing differential equation is given by: $$$$\left(D - 1\right)\left(D + 1\right)\left[ J_{n}(s)\right] = -J_{n - 1}(s)$$$$ Thus, $$$$J_n(s) = -\left(\left(D - 1\right)\left(D + 1\right)\right)^{-1}\left[ J_{n-1}(s)\right]$$$$ Which is my reasoning is correct implies that \begin{align} J_n(s) &= (-1)^n \left(\left(D - 1\right)\left(D + 1\right)\right)^{-(n - 1)}\left[ J_1(s)\right] = (-1)^n \left(\left(D - 1\right)\left(D + 1\right)\right)^{-(n - 1)}\left[ \frac{\pi}{2}e^{-s}\right] \nonumber \\ &= (-1)^n \frac{\pi}{2} \left(\left(D - 1\right)\left(D + 1\right)\right)^{-(n - 1)}\left[ e^{-s}\right] \end{align}

• Notation $J_n(.)$ is reserved for Bessel functions. Using that here may make some mistake, perhaps!!! Commented Jul 6, 2019 at 6:33
• Perhaps. It is defined in the post though.
– user679268
Commented Jul 6, 2019 at 6:47
• Using residue calculus, it is easy to obtain $$J_{n-1}=\pi e^{-s}\cdot\frac{n-1}{(-4)^n}\sum^n_{r=0}\binom{n}{r}\frac{(-2s)^r}{(r+1)!}$$ Commented Jul 6, 2019 at 10:04
• Sorry, I messed up with the signs. The correct answer should be $$J_{n+1}(s)=\frac{\pi e^{-s}}2\cdot\frac{n+1}{4^n}\sum^n_{r=0}\binom{n}{r}\frac{(2s)^r}{(r+1)!}$$ Commented Jul 6, 2019 at 10:21
• @Szeto - I very much appreciate your solution, but I was hoping to solve it using Real Analysis.
– user679268
Commented Jul 6, 2019 at 10:58

Start with the result (link): $$\int_{0}^{\infty }{\frac{\cos \left( su \right)}{\left( {{u}^{2}}+p \right)}du}=\frac{\pi {{e}^{-s\sqrt{p}}}}{2\sqrt{p}}$$ Differentiating both sides $$n-1$$ times (w.r.t $$p$$) $$\int_{0}^{\infty }{\frac{\left( n-1 \right)!{{\left( -1 \right)}^{n-1}}\cos \left( su \right)}{{{\left( {{u}^{2}}+p \right)}^{n}}}du}=\frac{{{d}^{n-1}}}{d{{p}^{n-1}}}\left( \frac{\pi {{e}^{-s\sqrt{p}}}}{2\sqrt{p}} \right)$$ Setting $$p=1$$ $$\int_{0}^{\infty }{\frac{\cos \left( su \right)}{{{\left( {{u}^{2}}+1 \right)}^{n}}}du}=\frac{1}{{{\left( -1 \right)}^{n-1}}\left( n-1 \right)!}{{\left[ \frac{{{d}^{n-1}}}{d{{p}^{n-1}}}\left( \frac{\pi {{e}^{-s\sqrt{p}}}}{2\sqrt{p}} \right) \right]}_{p=1}}$$ Note that the integral in question is indeed an integral representation(see equation 5 here) of the Modified Bessel Function of the Second Kind $${{K}_{n}}\left( s \right)$$ which is a solution to the Modified Bessel Differential Equation. After some research in special functions text-books I have found that almost (if not all authors) use Complex analysis methods to evaluate it , that’s why I strongly believe that forming a differential equation to find the integral is not an accessible method!!! and by the way here is the value of the integral in terms of speatial functions : $$\frac{\sqrt{\pi }{{2}^{\frac{1}{2}-n}}{{K}_{\frac{1}{2}-n}}\left( s \right)}{{{s}^{\frac{1}{2}-n}}\Gamma \left( n \right)}$$

• fantastic solution!
– user679268
Commented Jul 6, 2019 at 16:06
• thanks and have a nice day
– logo
Commented Jul 6, 2019 at 16:29
• I'm still upset with myself that I didn't see that simple pattern, but very happy you posted. I will have to keep this one in my 'toolkit'. Re Differential Equation - indeed, but hey with Leibniz's Integral rule differentiation under the curve works perfectly.
– user679268
Commented Jul 7, 2019 at 1:36

Recognize that the integral $$I_n(t) = \int_0^\infty \frac{\cos(xt)}{\left(x^2 + a^2\right)^n}\:dx$$ satisfies the recursion $$I_{n+1}(t)=\frac1{2n a^2}\left[(2n-1) I_{n}(t)-t I’_{n}(t)\right]$$ which can be evaluated successively as follows \begin{align} & I_1(t)=\int_{0}^{\infty }{\frac{\cos \left( xt \right)}{ {{x}^{2}}+a^2}dx}=\frac{\pi}{2a} {e}^{-at} \\ &I_2 (t)= \frac1{2 a^2}\left[ I_{1}(t)-t I’_{1}(t)\right] =\frac\pi{4a^3}e^{-at}(1+at)\\ &I_3 (t)= \frac1{4 a^2}\left[3 I_{2}(t)-t I’_{2}(t)\right] =\frac\pi{16 a^5}e^{-at}(3+3at + a^2t^2)\\ &I_4 (t)= \frac1{6 a^2}\left[5 I_{3}(t)-t I’_{3}(t)\right] =\frac\pi{96 a^7}e^{-at}(15+15at + 6a^2t^2+a^3t^3)\\ &I_5 (t)= \frac1{8 a^2}\left[7 I_{4}(t)-t I’_{4}(t)\right]=\cdots \end{align}