0-e-x-cos-x-2-dx-Discuss-the-convergence-of-this-generalised-intergral-Please-help-

Question Number 97563 by 675480065 last updated on 08/Jun/20

\int_{0}^{+ \infty} e^{- x} \cos (x^{2}) dx .

Discuss the convergence of this

generalised intergral .

Please help

Answered by mathmax by abdo last updated on 08/Jun/20

I = \int_{0}^{\infty} e^{- x} \cos (x^{2}) dx for all a > 0 x \to e^{- x} \cos (x^{2}) is continue on [0, a] so

integrable on [0, a] at [a, + \infty [\lim_{x \to + \infty} e^{- x} \cos (x^{2}) dx = 0 \Rightarrow

\int_{a}^{+ \infty} e^{- x} \cos (x^{2}) dx cv \Rightarrow \int_{0}^{\infty} e^{- x} \cos (x^{2}) dx converge .

the vslue I = Re (\int_{0}^{\infty} e^{- x + {ix}^{2}} dx) and

\int_{0}^{\infty} e^{- x + {ix}^{2}} dx = \int_{0}^{\infty} e^{{(\sqrt{i} x)}^{2} - 2 \sqrt{i} x \times \frac{1}{2 \sqrt{i}} + {(\frac{1}{2 \sqrt{i}})}^{2} - {(\frac{1}{2 \sqrt{i}})}^{2}} dx

= \int_{0}^{\infty} e^{{(\sqrt{i} x - \frac{1}{2 \sqrt{i}})}^{2} - \frac{1}{4 i}} dx = e^{\frac{i}{4}} \int_{0}^{\infty} e^{{(\sqrt{i} x - \frac{1}{2 \sqrt{i}})}^{2}} dx =_{\sqrt{i} x - \frac{1}{2 \sqrt{i}} = t} e^{\frac{i}{4}} \int_{- \frac{1}{2 \sqrt{i}}}^{+ \infty} e^{- t^{2}} \frac{dt}{\sqrt{i}}

= e^{- \frac{i π}{4}} e^{\frac{i}{4}} \int_{- \frac{1}{2} e^{- \frac{i π}{4}}}^{+ \infty} e^{- t^{2}} dt = e^{(- \frac{π}{4} + \frac{1}{4}) i} {\frac{\sqrt{π}}{2} - \int_{0}^{- \frac{1}{2} e^{- t^{2}}} dt} \dots be continued \dots

Commented by 675480065 last updated on 08/Jun/20

t h a n k s t e a c h e r

Commented by abdomathmax last updated on 08/Jun/20

you are welcome .