1. difficult Integration

Calculate

$\120dpi \int_{0}^{\infty }\frac{sinx}{1+x^{2}}dx$

2. Originally Posted by dapore
Calculate

$\120dpi \int_{0}^{\infty }\frac{sinx}{1+x^{2}}dx$
The only thing I can say is that your lovely integral is "absolute convergent" ...

It's hard to believe that someone sane will ask you to compute this...

3. It is "good" integral
at x=0 =0
at x=inf decreases as $1/x^2$
and its value is
integrate sin x dx&#47; &#40;1&#43;x&#94;2&#41; from x&#61;0 to &#43;inf - Wolfram|Alpha

4. Thank you I thought in the following way :
By integration by parts, we know that $2\int_{0}^{\infty}a e^{-a^{2}(1+x^{2})}da =\frac{1}{1+x^{2}}$

We then get:
$I =\int_{0}^{\infty}\frac{\sin x}{1+x^{2}}dx = 2\int_{0}^{\infty}ae^{-a^{2}}\boxed{\int_{0}^{\infty}e^{-a^{2}x^{2}}\sin x dx}da$
by working out the boxed part using integration by parts we get the answer

5. Originally Posted by dapore
by working out the boxed part using integration by parts we get the answer.
And how do you propose that we work out the boxed part? I would be interested to know.

6. Originally Posted by dapore
Calculate

$\120dpi \int_{0}^{\infty }\frac{sinx}{1+x^{2}}dx$
Use the Residue Theorem.

7. Who hasn't at disposal Wolfram's tools but only a table of Laplace Tranforms can remember that is...

$\displaystyle \varphi(s)= \mathcal{L} \{\frac{1}{1+t^{2}}\} = \int_{0}^{\infty} \frac{e^{-s\ t}}{1+t^{2}}\ dt = \cos s \ \{\frac{\pi}{2} - Si (s) \} - \sin s \ Ci(s)$ (1)

... so that is...

$\displaystyle \int_{0}^{\infty} \frac{\sin t}{1+t^{2}} \ dt = Im \{ \varphi(-i)\}$ (2)

Kind regards

$\chi$ $\sigma$

8. Wow! Chinese!