When I attempted trying to evaluate $$\int_{0}^{\infty} \frac{\sin^{2}{x}}{x(x^{2}+1)} dx$$, I first tried partial fractions to exploit the fact that an inverse tangent might be able to be found, which resulted in two integrals, the first one of which I highly suspect to be integrated using the Leibniz Integral rule, but I have no clue how to evaluate the second part which is $$\int_{0}^{\infty} \frac{x\sin^{2}{x}}{x^{2}+1} dx$$. I then attempted using infinite geometric series but it didn't really go anywhere. Is there something I'm missing here?
Asked
Active
Viewed 245 times
3
-
1Have you tried Feynman trick? – openspace Jun 19 '24 at 13:00
-
2See here. – Sahaj Jun 19 '24 at 13:00
-
@Sahaj that's not the same integral, but I wonder if there's a typo in the question... – Chris Lewis Jun 19 '24 at 13:06
-
Does it have a neat closed form? – Math-fun Jun 19 '24 at 13:07
-
1@Math-fun I put it into wolframalpha and it returned something with two exponential integrals, the Euler-mascheroni constant, and euler's number. I should think that is not too bad. – Jeremiah Tan Jun 19 '24 at 13:58
-
Wolfram Alpha says integral_0^∞ (sin^2(x))/(x (x^2 + 1)) dx = 1/4 (-(e^4 Ei(-2) + Ei(2))/e^2 + 2 gamma + log(4))≈0.557893 – marty cohen Jun 19 '24 at 17:04
1 Answers
2
$$I=\int\frac{\sin^{2}(x)}{x(x^{2}+1)}\, dx=\frac 12 \int\frac{1-\cos(2x)}{x(x^{2}+1)} \,dx$$ $$I=\frac 12\log (x)-\frac{1}{4} \log \left(x^2+1\right)-\frac 12\int\frac{\cos(2x)}{x(x^{2}+1)}\, dx$$
$$\frac{1}{x(x^{2}+1)}=\frac{1}{x(x+i)(x-i)}=\frac{1}{x}-\frac{1}{2 (x-i)}-\frac{1}{2 (x+i)}$$ So, we face three integrals $$J_a=\int \frac{\cos(2x)}{x+a}\, dx$$ $$x=t-a \quad \implies \quad J_a=\cos(2a)\int \frac{\cos(2t)}{t}\, dt+\sin(2a)\int \frac{\sin(2t)}{t}\, dt$$ and here appear the trigonometric integrals $$\int \frac{\cos(2t)}{t}\, dt=\text{Ci}(2 t)\qquad \text{and} \qquad \int \frac{\sin(2t)}{t}\, dt=\text{Si}(2 t)$$ which, for sure, are related to the exponential integral function (use Euler representation of the sine and cosine function).
Recombine all terms and simplify before using the limits and you will obtain Wolfram Alpha result or some other equivalent formulae.
Edit
The antiderivative is then given by $$8\,e^2\,I(x)=\left(1+e^4\right) \text{Ci}(2 i-2 x)-4 e^2 \text{Ci}(2 x)+e^4 \text{Ci}(2 x+2 i)+$$ $$\text{Ci}(2 x+2 i)+i e^4 \text{Si}(2 i-2 x)-i \text{Si}(2 i-2 x)+$$ $$i e^4 \text{Si}(2 x+2 i)-i \text{Si}(2 x+2 i)-2 e^2 \log \left(x^2+1\right)+4 e^2 \log (x)$$
Which gives $$8\,e^2\,I=-2 (\text{Shi}(2)+\text{Ci}(2 i))+e^4 (-2 \text{Ci}(2 i)+2 \text{Shi}(2)+i \pi )+i \pi +4 e^2 (\gamma +\log (2))$$ which is $$I=\frac{\gamma +\log (2)}{2}-\frac{e^4 \text{Ei}(-2)+\text{Ei}(2)}{4 e^2} $$
Claude Leibovici
- 289,558