- Thread starter idontknow
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Just eye-balling the thing, this should be doable using contour integration. Are you allowed to use this?Evaluate \(\displaystyle I=\int \limits_{0}^{\infty } \dfrac{\sin^2 (x)}{1+x^2 } dx\).

-Dan

Yes it is allowed.Just eye-balling the thing, this should be doable using contour integration. Are you allowed to use this?

-Dan

\(\displaystyle \int_0^{ \infty} \dfrac{sin^2(x)}{1 + x^2} ~ dx = \dfrac{1}{2} \int_{-\infty}^{ \infty} \dfrac{sin^2(x)}{1 + x^2} ~ dx\).

Now evaluate \(\displaystyle

\dfrac{1}{2} \int_{-\infty}^{ \infty} \dfrac{sin^2(z)}{1 + z^2} ~ dz\).

The poles of the integrand are at \(\displaystyle z = \pm i\). Choose a contour as the upper half circle going from -R to R and circling counterclockwise with radius R. We then let \(\displaystyle R \to \infty\). The contour encloses the \(\displaystyle z = i\) pole, which has a residue of \(\displaystyle - \dfrac{i}{2} sinh^2(1)\) at z = i so

\(\displaystyle \lim_{R \to \infty} \int_{-R}^{R} \dfrac{sin^2(z)}{1 + z^2} ~ dz + \lim_{R \to \infty} \int_{upper} \dfrac{sin^2(z)}{1 + z^2} ~ dz = 2 \pi i \left (- \dfrac{i}{2} sinh^2(1) \right )\)

The upper half circle integral goes to 0 in the limit as R goes to infinity, so we are left with

\(\displaystyle \int _0 ^{\infty} \dfrac{sin^2(x)}{1 + x^2} ~ dx = \lim_{R \to \infty} \dfrac{1}{2} \int_{-R}^{R} \dfrac{sin^2{z}}{1 + z^2} ~dz = \dfrac{ \pi }{2} sinh^2(1)\)

Which is wrong?? Where did I screw up?

-Dan

\(\displaystyle 2I=\int_{0}^{\infty} \frac{dt}{1+t^2 }dt -I_2 =\frac{\pi}{2}-I_2\) .

\(\displaystyle I_m '' =I_m \implies I_m = c_1 e^m +c_2 e^{-m}\) ; \(\displaystyle \; \; \begin{cases} c_1 +c_2 = I_0 =\pi /2 \\ c_1 e +c_2 /e =I_1 =\pi /2e \end{cases} \implies c_2 =\pi /2 , c_1 =0\).

\(\displaystyle I_m =e^{-m}\pi /2 \) ; \(\displaystyle \; \; I_2 =\frac{\pi}{2e^2}\) ; \(\displaystyle \; \; 2I=\pi /2 -\frac{\pi}{2e^2 }=\frac{\pi (e^2 -1)}{2e^2}\).

\(\displaystyle I=\frac{\pi (e^2 -1 )}{4e^2 }\).

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