How do I integrate this without contour integration?
$$\int_0^\infty \frac{x\sin(tx)}{1+x^2} dx$$
I have tried everything, splitting the integral from $0$ to $1$ and $1$ to infinity and using the geometric series summation, rewriting $\sin(tx)$ as $\frac{e^{itx}-e^{-itx}}{2i}$ and $1+x^2$ as $(x+i)(x-i)$.
But nothing has borne fruit. I am fine with using other well known functions and even complex analysis but I want to solve this without contour integration or other complicated methods like Laplace transforms.
Best Answer
Let'ss define a function $$ f(t)=\int_0^\infty \frac{x}{x^2+1} \sin(tx)dx$$ now let's use Laplace transfrom $$ L_t(f(t))(s)=L_t\left(\int_0^\infty \frac{x}{x^2+1} \sin(tx)dx\right)(s)$$ then $$ L_t(f(t))(s)=\int_0^\infty \frac{x}{x^2+1} L_t(\sin(tx))(s)dx=\int_0^\infty \frac{x}{x^2+1} \frac{x}{x^2+s^2} dx$$ and where $$ \frac{x}{x^2+1} \frac{x}{x^2+s^2}=\frac{1}{s^2-1} \frac{s^2}{x^2+s^2}-\frac{1}{s^2-1} \frac{1}{x^2+1} $$ So $$ L_t(f(t))(s)=\frac{1}{s^2-1} \frac{\pi}{2} s-\frac{1}{s^2-1} \frac{\pi}{2}=\frac{\pi}{2}\frac{1}{s+1} $$ and by taking the inverse Laplace transform we easily get $$ f(t)=\frac{\pi}{2} e^{-t} , Re(t)>0 $$ or $$ f(t)=\frac{\pi}{2} e^{-|t|} , t\in R $$
Addendum
If you don't know about Laplace transfrom we can rewrite the solution by using double integral instead of that trnasform because Laplace transform means integral.
Let's use the known integral $$ \int_0^\infty \sin(tx) e^{-st} dt=\frac{x}{x^2+s^2} $$ So $$ \int_0^\infty f(t) e^{-st} dt=\int_0^\infty \frac{x}{x^2+1} \frac{x}{x^2+s^2} dx$$ then it's easy to show that $$ \int_0^\infty f(t) e^{-st} dt=\frac{\pi}{2}\frac{1}{s+1} $$ but $$ \int_0^\infty \frac{\pi}{2}e^{-t} e^{-st} dt =\frac{\pi}{2}\frac{1}{s+1}$$ So we get (for positive $s$ or it's real part being positive if it's a complex number) $$ f(t)=\frac{\pi}{2}e^{-t} $$