24bingo sir.....
and yeah i ahve been visiting library often these days..[1]
now any taker for second one ?
1)\int_{0}^{\infty}{e^{-x^2}}dx=??
2) Given that \int_{0}^{\infty}{\frac {sinx}{x}}dx=\pi /2
findI=\int_{0}^{\infty}{\frac {sin^2x}{x^2}}dx
I know these all are impossbole integrals..but just found the soln to them toaday ....really nice...complex surely
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6 Answers
62Eureka you are reading a lot of stuff these days ;)
The first one is pretty well known... using a very very brilliant trick..
I will give the first hint ;)
\\I=\int_{0}^{\infty}e^{-x^2}dx \\I=\int_{0}^{\infty}e^{-y^2}dy \\I^2=(\int_{0}^{\infty}e^{-y^2}dy)(\int_{0}^{\infty}e^{-x^2}dx) \\I^2=\int_{0}^{\infty}\int_{0}^{\infty}e^{-x^2-y^2}dxdy
Now try the subtle "complex" hint that eureka has just given :D
24
66Q2) Use parametric integration: Consider, instead,
I(a)=\int_0^\infty \dfrac{\sin^2ax}{x^2}\ \mathrm dx
Differentiating w.r.t. a, we get
I^\prime(a)=\int_0^\infty \dfrac{2x\sin ax\cos ax}{x^2}\ \mathrm dx = \int_0^\infty \dfrac{\sin2 ax}{x}\ \mathrm dx
Next, substitute 2ax = y, then dx = dy2a. So we get
I^\prime(a)=\int_0^\infty \dfrac{\sin y}{y}\ \mathrm dy = \dfrac{\pi}{2}
And so
I(a)= \dfrac{a\pi}{2}+C
Since I(0)=0, so C=0. Hence,
I(a)= \dfrac{a\pi}{2}
Thus,
\int_0^\infty \dfrac{\sin^2x}{x^2}\ \mathrm dx = I(1)= \dfrac{\pi}{2}