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This question already has an answer here:

$$\int_0^{\infty} \frac{\sin^2(x)}{x^2(x^2+1)} dx$$

The integral is equals with $\frac{\pi}{4}+\frac{\pi}{4e^2}$, but i can't prove it.

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marked as duplicate by user186170, tired, user91500, Claude Leibovici, Crostul Oct 29 '15 at 15:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Let $$I(a)=\int_{0}^{\infty}\frac{\sin^2 ax}{x^2(x^2+1)}dx=\int_0^\infty \frac{\sin^2 ax}{x^2}dx-\int_0^\infty \frac{\sin^2ax}{(1+x^2)}dx\\=\frac{\pi a}{2}-\int_0^\infty\frac{\sin^2 ax}{1+x^2}dx$$ Here I have used the result $$\int_{0}^\infty \frac{\sin^2 x}{x^2}dx=\pi/2$$ Then, $$dI/da=\int_{0}^\infty \frac{\sin 2ax}{x(x^2+1)}\\\implies d^2I/da^2=2\int_{0}^\infty \frac{\cos 2ax}{x^2+1}dx\\=2\int_{0}^\infty\frac{1-2\sin^2ax}{1+x^2}dx\\=2\pi/2-4\int_{0}^\infty\frac{\sin^2 ax}{1+x^2}dx\\=\pi-4(a\pi/2-I(a))\\\implies d^2I/da^2=4I-2a\pi+\pi$$ The CF is $$C_1 e^{2a}+C_2e^{-2a}$$ and the PI is $$\pi\frac{1}{D^2-4}(1-2a)=\frac{1}{D-2}\pi e^{-2a}\int (1-2a)e^{2a}da\\=\frac{1}{D-2}\pi e^{-2a}(1/2 e^{2a}-ae^{2a}+1/2e^{2a})=\frac{1}{D-2}\pi (1-a)\\=\pi e^{2a}\int e^{-2a}(1-a)da=\pi(-1/2e^{-2a}+a/2e^{-2a}+1/4e^{-2a})e^{2a}\\=\pi(a/2-1/4)$$ So, $$I(a)=C_1e^{2a}+C_2e^{-2a}+\pi(a/2-1/4)$$Now, $$I(0)=0, dI(0)/da=0\\\implies C_1+C_2=\pi/4\\2a(C_1-C_2)=-\pi/2\\\implies C_1=\pi/8-\pi/8a, \ C_2=\pi/8+\pi/8a$$ Hence, the desired integral is $$I(1)=C_1(a=1)e^2+C_2(a=1)e^{-2}+\pi/4=\pi/4+\pi/4e^2$$

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  • $\begingroup$ How do you get that d^2I/da^2 equals with pi-4*that integrate? $\endgroup$ – Botond Oct 30 '15 at 8:26
  • $\begingroup$ I will edit the answer to make that point clear. $\endgroup$ – Samrat Mukhopadhyay Oct 31 '15 at 5:55

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