Интеграл cos(x*x) (dx)

Вы ввели

  1            
  /            
 |             
 |  cos(x*x) dx
 |             
/              
0              

$$\int_{0}^{1} \cos{\left (x x \right )}\, dx$$

Ответ

                                       /  ___ \           
                    ___   ____         |\/ 2  |           
  1               \/ 2 *\/ pi *fresnelc|------|*gamma(1/4)
  /                                    |  ____|           
 |                                     \\/ pi /           
 |  cos(x*x) dx = ----------------------------------------
 |                              8*gamma(5/4)              
/                                                         
0                                                         

$$-{{\sqrt{\pi}\,\left(\left(\sqrt{2}\,i-\sqrt{2}\right)\, \mathrm{erf}\left({{\sqrt{2}\,i+\sqrt{2}}\over{2}}\right)+\left( \sqrt{2}\,i+\sqrt{2}\right)\,\mathrm{erf}\left({{\sqrt{2}\,i-\sqrt{2 }}\over{2}}\right)+\left(-\sqrt{2}\,i-\sqrt{2}\right)\,\mathrm{erf} \left(\sqrt{-i}\right)+\left(\sqrt{2}\,i-\sqrt{2}\right)\, \mathrm{erf}\left(\left(-1\right)^{{{1}\over{4}}}\right)\right) }\over{16}}$$

Численный ответ

0.904524237900272

Ответ (Неопределённый)

                                          /    ___\           
                       ___   ____         |x*\/ 2 |           
                     \/ 2 *\/ pi *fresnelc|-------|*gamma(1/4)
  /                                       |   ____|           
 |                                        \ \/ pi /           
 | cos(x*x) dx = C + -----------------------------------------
 |                                  8*gamma(5/4)              
/                                                             

$$-{{\sqrt{\pi}\,\left(\left(\sqrt{2}\,i-\sqrt{2}\right)\, \mathrm{erf}\left({{\left(\sqrt{2}\,i+\sqrt{2}\right)\,x}\over{2}} \right)+\left(\sqrt{2}\,i+\sqrt{2}\right)\,\mathrm{erf}\left({{ \left(\sqrt{2}\,i-\sqrt{2}\right)\,x}\over{2}}\right)+\left(-\sqrt{2 }\,i-\sqrt{2}\right)\,\mathrm{erf}\left(\sqrt{-i}\,x\right)+\left( \sqrt{2}\,i-\sqrt{2}\right)\,\mathrm{erf}\left(\left(-1\right)^{{{1 }\over{4}}}\,x\right)\right)}\over{16}}$$