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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.

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  • $\begingroup$ Probably Feynman trick should work, have you tried it? $\endgroup$
    – Zima
    Commented Feb 9 at 9:49
  • $\begingroup$ 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 $\endgroup$
    – uggupuggu
    Commented Feb 9 at 9:51
  • $\begingroup$ I assumed $t>0$ in my answer $\endgroup$ Commented Feb 9 at 10:35

2 Answers 2

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

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  • $\begingroup$ Im not really familiar with laplace transforms... ill add that to my question $\endgroup$
    – uggupuggu
    Commented Feb 9 at 9:58
  • $\begingroup$ @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) $\endgroup$
    – Faoler
    Commented Feb 9 at 20:41
  • $\begingroup$ I decided to suck it up and learn what laplace transforms are, thanks $\endgroup$
    – uggupuggu
    Commented Feb 14 at 13:13
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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 $$I_n=\int_z^\infty \frac{x\sin^{2n+1}(tx)}{1+x^2}\, dx$$

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

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