Question Number 131049 by mathmax by abdo last updated on 31/Jan/21
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{xt}\right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\:\mathrm{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$
Answered by mindispower last updated on 01/Feb/21
$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({xt}\right)}{{x}^{\mathrm{2}} +{t}^{\mathrm{2}} }{dt} \\ $$$${t}={xr}\Rightarrow{dt}={xdr} \\ $$$$=\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left({r}\right)}{\mathrm{1}+{r}^{\mathrm{2}} }=\frac{\mathrm{1}}{{x}}.\frac{\mathrm{1}}{\mathrm{2}}{Re}\int_{−\infty} ^{\infty} \frac{{e}^{{ir}} }{\mathrm{1}+{r}^{\mathrm{2}} }{dr} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{x}}{Res}\left(\frac{{e}^{{ir}} }{\mathrm{1}+{r}^{\mathrm{2}} },{r}={i}\right)=\frac{{i}\pi}{{x}},\frac{{e}^{−\mathrm{1}} }{\mathrm{2}{i}}=\frac{\pi}{{xe}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx},{not}\:{exist} \\ $$