Bourbaki

Exercise 35, § 1

Functions of a Real Variable, III, 124

By differentiating with respect to the parameter and using exerc. 33 c), prove the formulae

∫0+∞e−x2cos⁡αx dx=π2e−α2, \int_0^{+\infty} e^{-x^2} \cos \alpha x \, dx = \frac{\sqrt{\pi}}{2} e^{-\alpha^2},
∫0+∞1−e−αx2x2dx=πα \int_0^{+\infty} \frac{1 - e^{-\alpha x^2}}{x^2} dx = \sqrt{\pi \alpha}
∫0+∞exp⁡(−x2−α2x2)dx=π2e−2α(α>0). \int_0^{+\infty} \exp \left( -x^2 - \frac{\alpha^2}{x^2} \right) dx = \frac{\sqrt{\pi}}{2} e^{-2\alpha} \quad (\alpha > 0).

Solution

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Transcribed from the printed volume. Not translated.