# Tag Info

## New answers tagged definite-integrals

1

Here are closed forms for the first and second integrals respectively $$I_1 = { \frac {\sqrt {2\pi }}{b\sqrt {T}}}{{\rm e}^{-\,{\frac {{b}^{2}}{2T}}}}$$ $$I_2= \, \pi- {{\rm erf}\left( {\frac {b}{\sqrt {2T}}}\right)}.$$ You can test them numerically.

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From the hint, $\cos^2 x= \frac{1}{2}(\cos 2x+1)$ so our integral becomes $$\int_{-\pi/2}^{\pi/2} \frac{1}{2}(\cos 2x+1)dx= \left. \left (\frac{1}{4}\sin 2x+\frac{1}{2}x \right ) \right |_{-\pi/2}^{\pi/2}=\pi/2$$

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Consider $$\mathcal{I}(y,t)=\int_{-\infty}^{\infty}\frac{\cos xt}{x^2+y^2}\ dx=\frac{\pi e^{-yt}}{y}\quad;\quad\text{for}\ t>0.\tag1$$ Differentiating $(1)$ with respect $t$ and $y$ yields \begin{align} \frac{\partial^2\mathcal{I}}{\partial y\partial t}=\int_{-\infty}^{\infty}\frac{2xy\sin xt}{(x^2+y^2)^2}\ dx&=\pi te^{-yt}\\ ...

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