Integration and Summation Together !

Let $\displaystyle I(a,b)=\int_{0}^{\infty}\frac{\sin(ax)}{x(x^{2}+b)}dx$

So we have :-

$\displaystyle \frac{\partial I}{\partial a} = \int_{0}^{\infty}\frac{\cos(ax)}{x^{2}+b} dx$

$\displaystyle \frac{\partial^{2}I}{\partial a^{2}} = -\int_{0}^{\infty}\frac{x\sin(ax)}{x^{2}+b}dx= -\int_{0}^{\infty} \frac{\sin(ax)}{x}dx + b\int_{0}^{\infty}\frac{\sin(ax)}{x(x^{2}+b)} dx = -\frac{\pi}{2} + bI(a,b)$

Here I used the fact that $\displaystyle \int_{0}^{\infty}\frac{\sin(ax)}{x} = \frac{\pi}{2}$

So we have $\displaystyle (D^{2}-b)I = \frac{-\pi}{2}$ Here $\displaystyle D\equiv \frac{\partial}{\partial a}$ and $\displaystyle D^{2}\equiv \frac{\partial^{2}}{\partial a^{2}}$

Solving this ODE in a we get

$\displaystyle I(a,b) =C_{1}e^{a\sqrt{b}} + C_{2}e^{-a\sqrt{b}} +\frac{\pi}{2b}$ We have $I(0,b) = 0$

So $\displaystyle C_{1}+C_{2} + \frac{\pi}{2b} = 0$

Also $\displaystyle \frac{\partial I}{\partial a} = \sqrt{b}(C_{1}e^{a\sqrt{b}} -C_{2}e^{-a\sqrt{b}})$

Also $\displaystyle \frac{\partial I}{\partial a}$ at $(0,b)=$ $\displaystyle \int_{0}^{\infty}\frac{dx}{x^{2}+b} = \frac{\pi}{2\sqrt{b}}$ .

So we get $\displaystyle C_{1}=0$ and $\displaystyle C_{2} = \frac{-\pi}{2b}$

So $\displaystyle I(a,b) = \frac{-\pi}{2b}e^{-a\sqrt{b}} + \frac{\pi}{2b}$

$\displaystyle \frac{\partial I}{\partial a } = \frac{\pi}{2\sqrt{b}}e^{-a\sqrt{b}}$

$\displaystyle \frac{\partial^{2} I}{\partial b \partial a} = \frac{-\pi}{4}\left(\frac{ae^{-a\sqrt{b}}}{b} +\frac{e^{-a\sqrt{b}}}{b\sqrt{b}}\right)$

From the Definition of $I(a,b)$ we have

$\displaystyle \frac{\partial^{2} I}{\partial b \partial a} = \int_{0}^{\infty} \frac{-\cos(ax)}{(x^{2}+b)^{2}}dx$

We know $\displaystyle \int_{-\infty}^{\infty} \frac{\cos(nx)}{(x^{2}+1)^{2}} dx = 2\int_{0}^{\infty}\frac{\cos(nx)}{(x^{2}+1)^{2}}dx$

So what we need is $\displaystyle -2\frac{\partial^{2} I}{\partial b \partial a}$ at the point $(n,1)$ .

So Using the expression above we have $\displaystyle I_{n} = \frac{\pi}{2}(n+1)e^{-n}$ .

We know $\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty}x^{n}$ for $-1<x<1$

Hence $\displaystyle \frac{x}{1-x} = \sum_{n=0}^{\infty}x^{n+1}$

Differentiating term by term once on both sides (possible due to uniform convergence)

We have $\displaystyle \frac{1}{(1-x)^{2}} = \sum_{n=0}^{\infty}(n+1)x^{n}$

Using the above we have $\displaystyle \sum_{n=0}^{\infty}I_{n} =A= \frac{\pi}{2}\cdot \frac{e^{2}}{(e-1)^{2}} = 3.931159$

Hence $\displaystyle \lfloor 100A \rfloor =391$

Note 1:- The function $\frac{sin(ax)}{x(x^{2}+b)}$ is continuous for all x and all a and b>0. The partial derivative to it is continuous for all x and a and b>0. The improper integral of the function is uniformly convergent for all a by M-test .The improper integral of the partial derivative of the function is uniformly convergent for all positive $a$ .Hence justification of interchange of derivative and integral sign is valid as these form a sufficient set of conditions for us to do so.

Note 2:- I have used the operator method for finding the solution to the ordinary differential equation in terms of $a$ using auxiliary equations,complimentary functions and particular integrals. These are pretty standard things and can be found in any elementary level differential equations book.

Usually People Use Residue Theorem to solve this kind of problem but I did it using Feynmann Technique and found it very interesting as it leads to many generalizations. Please upvote if you liked the problem and the solution !.

We can solve $\displaystyle I_n = \int_{-\infty}^\infty \frac {e^{inx}}{(1+x^2)^2} dx = \Re \left(\int_{-\infty}^\infty \frac {\cos (nx)}{(1+x^2)^2} dx \right)$ using contour integration with a semicircle of radius $R \to \infty$ in the upper complex plain.

$\oint \frac {e^{inz}}{(1+z^2)^2} dz = \int_{-\infty}^\infty \frac {e^{inx}}{(1+x^2)^2} dx + \int_{\text{arc}} \frac {e^{inz}}{(1+z^2)^2} dz$

By Jordan's lemma, the last integral in the RHS above approaches $0$ as $R \to \infty$ . Then

$\begin{aligned} \int_{-\infty}^\infty \frac {e^{inx}}{(1+x^2)^2} dx & = \oint \frac {e^{inz}}{(1+z^2)^2} dz \\ & = \oint \frac {e^{inz}}{((z-i)(z+i))^2} dz & \small \blue{\text{Only 1 pole }z=i \text{ is within the contour.}} \\ & = 2\pi i\cdot \frac d{dz} \left[\frac {e^{inz}}{(z+i)^2} \right]_{z=i} & \small \blue{\text{By Cauchy integral formula}} \\ & = 2\pi i \left[\frac {e^{iz}(inz-n-2)}{(z+i)^3} \right]_{z=i} \\ & = \frac {\pi e^{-n}(n+1)}2 \end{aligned}$

$\displaystyle I_n = \Re \left( \int_{-\infty}^\infty \frac {e^{inx}}{(1+x^2)^2} dx \right) = \frac {\pi e^{-n}(n+1)}2$ .

Then

$\begin{aligned} A & = \sum_{n=0}^\infty \frac {\pi e^{-n}(n+1)}2 = \frac \pi 2 \sum_{n=1}^\infty \frac {d x^n}{dx} \bigg|_{x = e^{-1}} = \frac \pi 2 \cdot \frac d{dx} \sum_{n=1}^\infty x^n \bigg|_{x = e^{-1}} = \frac \pi 2 \cdot \frac d{dx} \left[\frac x{1-x} \right]_{x = e^{-1}} = \frac {\pi e^2}{2(e-1)^2} \approx 3.931 \end{aligned}$

Therefore $\lfloor 100A\rfloor = \boxed{393}$ .

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Reference: Cauchy integral formula