advnced-calculus-prove-0-sin-tan-x-x-dx-pi-2-1-1-e-prove-that-0-1-e-x-2-x-2-dx-pi-

Question Number 133490 by mnjuly1970 last updated on 22/Feb/21

\dots . a d v n c e d c a l c u l u s \dots .

p r o v e : ϕ = \int_{0}^{\infty} \frac{s i n (t a n (x))}{x} d x = \frac{π}{2} (1 - \frac{1}{e})

p r o v e t h a t : Φ = \int_{0}^{\infty} \frac{1 - e^{- x^{2}}}{x^{2}} d x = \sqrt{π}

Answered by Dwaipayan Shikari last updated on 22/Feb/21

I (α) = \int_{0}^{\infty} \frac{e^{- α x^{2}} - e^{- β x^{2}}}{x^{2}} d x

I^{'} (α) = - \int_{0}^{\infty} e^{- α x^{2}} d x = - \sqrt{\frac{π}{4 α}} \Rightarrow I (α) = - \sqrt{π α} + C

I (β) = - \sqrt{π β} + C = 0 \Rightarrow C = \sqrt{π β}

I (α) = \sqrt{π} (\sqrt{β} - \sqrt{α})

α = 0 β = 1

I (0) = \int_{0}^{\infty} \frac{1 - e^{- x^{2}}}{x^{2}} d x = \sqrt{π}

Commented by mnjuly1970 last updated on 22/Feb/21

g r a t e f u l . . m r p a y a n

v e r y n i c e . .

Answered by mathmax by abdo last updated on 22/Feb/21

Φ = \int_{0}^{\infty} \frac{1 - e^{- x^{2}}}{x^{2}} dx let f (a) = \int_{0}^{\infty} \frac{1 - e^{- {ax}^{2}}}{x^{2}} with a > 0

f^{^{'}} (a) = \int_{0}^{\infty} e^{- {ax}^{2}} dx = \int_{0}^{\infty} e^{- {(\sqrt{a} x)}^{2}} dx =_{\sqrt{a} x = y} \int_{0}^{\infty} e^{- y^{2}} \frac{dy}{\sqrt{a}}

= \frac{1}{\sqrt{a}} \int_{0}^{\infty} e^{- y^{2}} dy = \frac{1}{\sqrt{a}} \frac{\sqrt{π}}{2} \Rightarrow f (a) = k + \sqrt{π a}

\lim_{a \to o} f (a) = 0 \Rightarrow f (a) = \sqrt{π a} and Φ = f (1) = \sqrt{π}