Integrating $\int_0^\infty \frac{x\sin(tx)}{1+x^2} dx$ without contour integration?

Question

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.

Answer 1

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) $$ f(t)=\frac{\pi}{2}e^{-t} $$

Answer 2

Start with $$\frac{x}{1+x^2} =\frac{x}{(x+i)(x-i)}=\frac{1}{2 (x+i)}+\frac{1}{2 (x-i)}$$ Now consider $$I_+=\int \frac{\sin(tx)}{x+i}\,dx=\int \frac{\sin (t (y-i))}{y}\,dy$$ $$\sin (t (y-i))=\cosh (t) \sin (t y)-i \sinh (t) \cos (t y)$$ $$\int \frac{\sin (ty)}{y}\,dy=t \int \frac{\sin (u)}{u}\,du=t\,\text{Si}(u)$$ $$\int \frac{\cos (ty)}{y}\,dy=t \int \frac{\cos (u)}{u}\,du=t\,\text{Ci}(u)$$

So, we have all required antiderivatives.

Back to $x$ $$2\int \frac{x\sin(tx)}{1+x^2} dx=i \sinh (t) (\text{Ci}(-t (x-i))-\text{Ci}(t (x+i)))+$$ $$\cosh (t) (\text{Si}(t (x+i))-\text{Si}(i t-t x))$$ which is $0$ for $x=0$ $$\int_0^\infty \frac{x\sin(tx)}{1+x^2}\, dx=\frac{ \pi}{2} (\cosh (t)-\sinh (t))= \frac \pi 2\, e^{-t}$$

Edit

We can even go further using asymptotics $$\int_0^z \frac{x\sin(tx)}{1+x^2}\, dx= \frac \pi 2\, e^{-t}-\frac{\cos (t z)}{t z} A-\frac{\sin (t z)}{(t z)^2}B$$ with $$A=1-\frac{t^2+2}{(tz)^2}+\frac{t^4+12 t^2+24}{(tz)^4 }-\frac{t^6+30 t^4+360 t^2+720}{(tz)^6 }+O\left(\frac{1}{z^8}\right)$$ $$B=1-\frac{3 \left(t^2+2\right)}{(tz)^2}+\frac{5 (t^4+10 t^2+24) }{(tz)^4}-\frac{7(t^6+30 t^4+360 t^2+720)}{(tz)^6}+O\left(\frac{1}{z^8}\right)$$

Computing the value of $$\int_z^\infty \frac{x\sin(tx)}{1+x^2}\, dx$$ for $t=3$ and $z=10$, the exact value is $4.0212430\times 10^{-3}$ while the above asymptotics gives $4.0212455\times 10^{-3}$ (absolute error of $2.56\times 10^{-9}$).

We can easily generalize the problem to $$I_n=\int_z^\infty \frac{x\sin^{2n+1}(tx)}{1+x^2}\, dx$$