# Help in two improper integrals

For a physics problem, I need to evaluate two integrals: $$I_n=\int_{0}^{\infty} f(x) dx=\int_{0}^{\infty} \frac{x^2 \sin^2x}{(n^2\pi^2-x^2)^2} dx,$$ and $$J_n=\int_{0}^{\infty} \frac{x^2 \cos^2x}{((n+1/2)^2\pi^2-x^2)^2} dx, n=1,2,3..$$ At Mathematica the value of these integral turns out to be $$\pi/4$$ which is independent of $$n$$.

$$I_n$$ has got a pole of order 2 at $$x=n\pi$$ and when I calculate the residue at $$x=n \pi$$ is zero. But on the other hand I find that $$\lim_{x \to n\pi} f(x)=0,$$ so $$x=n\pi$$ may not be a pole of $$f(x)$$. I face the same problem in the case of $$J_n$$. Also $$I_0=\pi/2$$ is a standard integral. In any case, I want to evaluate both $$I_n$$ and $$J_n$$. Please help me.

• Try to use feynman technique – Aditya Dwivedi Jan 26 at 8:13
• Indeed $x=n \pi$ is a removable pole. – Crostul Jan 26 at 8:17

$$I_n=\int_{0}^{\infty} \frac{x^2 \sin^2x}{(n^2\pi^2-x^2)^2} dx$$. Integrating by part end expanding integration to the whole axis (the integrand is even) we get

$$I_n=-\frac{1}{2}\int_{0}^{\infty} \frac{\sin^2x+x\sin2x}{(n^2\pi^2-x^2)} dx=-\frac{1}{4}\int_{-\infty}^{\infty} \frac{x\sin2x}{(n^2\pi^2-x^2)}dx-\frac{1}{8}\int_{-\infty}^{\infty} \frac{1-\cos2x}{(n^2\pi^2-x^2)}dx=I_1+I_2$$.

Then we choose the closed contour $$C$$ in upper half of the complex plane:

To close the contour, we added two small half-circles $$C_1$$ and $$C_2$$ of radius $$r$$ - around $$x=-n\pi$$ and $$x=n\pi$$ $$(r\to0)$$, and a big circle of radius $$R$$ ($$R\to\infty$$). The required integral $$I_n$$ is equal to the principal value of the integral along the axis. Then, $$\sin2x=Im \exp(i2x)$$ and $$\cos2x=Re\exp(2ix)$$.

The integral over the closed contour $$\oint_C=0$$ (no singular points inside the contour), therefore$$I_n+I_{C1}+I_{C2}+I_R=0$$. It is easy to evaluate that $$I_R\to0$$ at $$R\to\infty$$ (using Jordan's lemma), so $$I_n=-I_{C1}-I_{C2}$$

$$I_1=-\frac{1}{4}Im\int_{-\infty}^{\infty} \frac{xe^{2ix}}{(n^2\pi^2-x^2)}dx=Im(-I_{C1}-I_{C2})=\frac{1}{4}Im\left((-\pi{i})[\frac{-\pi{n}\exp(-2\pi{in})}{2\pi{n}}+\frac{\pi{n}\exp(2\pi{in})}{-2\pi{n}}]\right)$$

$$I_1=\frac{\pi}{4}$$

Making similar calculations it is easy to show that $$I_2=0$$ Finally, $$I_n=I_1+I_2=\frac{\pi}{4}$$

$$J_n$$ can be evaluated in the same way.

Let is use $$\int_{n\pi}^{\infty} \frac{\sin^2 t}{t^2} dt=\frac{\pi}{2}-\text{Ei}(2n \pi)$$ Then $$I_n=\int_{0}^{\infty} \frac{x^2 \sin^2 x}{x^2-n^2\pi^2} dx$$ $$=\frac{1}{4}\int_{0}^{\infty} dx \left(\frac{\sin^2 x}{(x-n\pi)^2}+\frac{\sin^2 x}{(x+n\pi)^2}+\frac{\sin^2 x}{4n\pi(x-n\pi)}-\frac{\sin^2 x}{4n\pi(x+n\pi)}\right)$$ $$I_n=\frac{1}{4}[\pi/2-\text{Ei}(2n\pi)+\pi/2+\text{Ei}(2n\pi)]+\int_{-n\pi}^{n\pi} \frac{\sin^2 t}{t}dt=\frac{\pi}{4}$$ Similarly $$J_n=\int_{0}^{\infty} \frac{x^2 \cos^2 x}{x^2-m^2\pi^2} dx, m=n+1/2$$ $$=\frac{1}{4}\int_{0}^{\infty} dx \left(\frac{\cos^2 x}{(x-m\pi)^2}+\frac{\cos^2 x}{(x+m\pi)^2}+\frac{\cos^2 x}{4m\pi(x-m\pi)}-\frac{\cos^2 x}{4m\pi(x+m\pi)}\right)$$ $$I_n=\frac{1}{4}[\pi/2-\text{Ei}(2m\pi)+\pi/2+\text{Ei}(2m\pi)]+\int_{-m\pi}^{m\pi} \frac{\sin^2 t}{t}dt=\frac{\pi}{4}$$