advanced-calculus-prove-that-0-cos-2pix-2-cosh-2-pix-dx-1-4-

Question Number 141649 by mnjuly1970 last updated on 21/May/21

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

p r o v e t h a t - ::

ϕ := \int_{0}^{\infty} \frac{c o s (2 π x^{2})}{{c o s h}^{2} (π x)} d x = \frac{1}{4} ✓

Answered by ArielVyny last updated on 22/May/21

ϕ = \int_{0}^{\propto} \frac{c o s (2 π x^{2}) + 1}{{c o s h}^{2} (π x)} d x - \int_{0}^{\propto} \frac{1}{{c o s h}^{2} (π x)} d x

ϕ = ϕ_{1} - ϕ_{2}

ϕ_{2} = \int_{0}^{\propto} \frac{1}{{c o s h}^{2} (π x)} d x

π x = t π d x = d t

ϕ_{2} = \frac{1}{π} \int_{0}^{\propto} \frac{1}{{c o s h}^{2} (t)} d t = \frac{1}{π}

ϕ_{1} = \int_{0}^{\propto} \frac{c o s (2 π x^{2}) + 1}{{c o s h}^{2} (π x)} d x

{c o s}^{2} (π x^{2}) = \frac{1 + c o s (2 π x^{2})}{2}

ϕ_{1} = \int_{0}^{\propto} \frac{2 {c o s}^{2} (π x^{2})}{{c o s h}^{2} (π x)} d x

ϕ_{1} = 2 \int_{0}^{\propto} {(\frac{c o s (π x^{2})}{c o s h (π x)})}^{2} d x

ϕ_{1} = 2 \int_{0}^{\propto} {(\frac{{(- 1)}^{x^{2}}}{c o s h (π x)})}^{2} d x

ϕ_{1} = 2 \int_{0}^{\propto} {(\frac{1}{c o s h (π x)})}^{2} d x

ϕ_{1} = 2 \int_{0}^{\propto} \frac{1}{{c o s h}^{2} (π x} d x

π x = t . π d x = d t

ϕ_{1} = 2 \frac{1}{π} \int_{0}^{\propto} \frac{1}{{c o s h}^{2} (t)} d t

ϕ_{1} = \frac{2}{π}

ϕ = \frac{1}{π}

c h e c k i n t h e a n s w e r b e c a u s e i d o n o t h a v e y o u r r e s u l t .

Commented by mnjuly1970 last updated on 22/May/21

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