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

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.

• Probably Feynman trick should work, have you tried it?
– Zima
Commented Feb 9 at 9:49
• I would then get cos(tx)x^2/(1+x^2),I would then while solving get cos(tx)/(1+x^2) which is a well-known integral i suppose but it also involves contour integration Commented Feb 9 at 9:51
• I assumed $t>0$ in my answer Commented Feb 9 at 10:35

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 $$f(t)=\frac{\pi}{2} e^{-|t|} , t\in R$$

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

• Im not really familiar with laplace transforms... ill add that to my question Commented Feb 9 at 9:58
• @uggupuggu you should know the ways for solving this integral which are : Laplace transform (as I solved) and inverse laplace transform and by complex analysis and Fynman's trick which give us ODE and by special functions Ci(x) , Si(x) Commented Feb 9 at 20:41
• I decided to suck it up and learn what laplace transforms are, thanks Commented Feb 14 at 13:13

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 $$I_n=\int_z^\infty \frac{x\sin^{2n+1}(tx)}{1+x^2}\, dx$$

• Ci(−t(x−i)) How did you get this? Im trying to work it out Commented Feb 9 at 14:52
• @uggupuggu. Cosine integral function Commented Feb 9 at 15:08
• The formula in the edit is exact or an asymptotic approximation
– Zima
Commented Feb 9 at 17:52
• @Zima asymptotics for sure. Cheers Commented Feb 9 at 18:04
• @Zima. Asymptotics improved and tested Commented Feb 10 at 4:31