Vtyjkyuk

I found the following result (I apologize if this has been posted before, but I could not find anything here). I am wondering whether there is an approach without using contour integration.

Let $k$ be a nonnegative real number. Prove that $$int_0^infty,fracsin(kx)x,left(x^2+1right),textdx=fracpi2,Big(1-exp(-k)Big)=pi,expleft(-frack2right),sinhleft(frack2right),.$$

I am especially interested in a solution not using complex analysis. However, a complex-analytic solution that is different from mine is very welcome as well. I have three approaches, but all of them end up using complex analysis as a major part.

We also have this very nice consequence. This result can be proven on its own without knowing the integral $displaystyle int_0^infty,fracsin(kx)x,left(x^2+1right),textdx$.

Let $k$ be a nonnegative real number. Then,
$$int_0^infty,frac1-cos(kx)x^2,left(x^2+1right),textdx=fracpi2,big(k-1+exp(-k)big),.$$
Equivalently,
$$int_0^infty,fracsin^2(kx)x^2,left(x^2+1right),textdx=fracpi4,big(2k-1+exp(-2k)big),.$$

Interestingly, during my quest to obtain this integral, I discovered two more integral relations, although I do not know how to obtain the exact values of any of them. The exact values of these integrals involve the exponential integral $textEi$, where $textEi(x)=displaystyletextPVint_-infty^x,fracexp(t)t,textdt$ for all $xinmathbbR$. Both of the results below also came from contour integrations.

Let $k$ be a nonnegative real number. Then,
$$int_0^infty,frac1-cos(kx)x,left(x^2+1right),textdx=-frac2pi,int_0^infty,fracln(x),sin(kx)x,left(x^2+1right),textdx,.$$
and
$$int_0^infty,fracsin(kx)x^2+1,textdx=-frac2pi,int_0^infty,fracln(x),cos(kx)x^2+1,textdx,.$$

Mathematica says that
$$int_0^infty,fracsin(kx)x^2+1,textdx=fracexp(-k),textEi(+k)-exp(+k),textEi(-k)2$$
and that
$$int_0^infty,frac1-cos(kx)x,left(x^2+1right),textdx=gamma+ln(k)-fracexp(-k),textEi(+k)+exp(+k),textEi(-k)2,.$$
Here, $gammaapprox 0.57722$ is the Euler-Mascheroni constant.

Approach I.

Consider the meromorphic function $f(z):=dfracexp(textikz)z(z^2+1)$ for all $zin mathbbCsetminus 0,-texti,+texti$. For $epsilonin(0,1)$, let $C_epsilon$ be the positively oriented contour $$left[+epsilon,+frac1epsilonright]cup Biggl,thetain[0,pi]Biggrcupleft[-frac1epsilon,-epsilonright]cup Big,thetain[pi,0]Big,.$$ Write $Gamma_r$ for the positively-oriented (with respect to $0$) semicircle $Bigr,exp(textitheta),Big$ for every $r>0$. We have $$I(k):=limlimits_epsilonto 0^+,oint_C_epsilon,f(z),textdz=2pitexti,textRes_z=textibig(f(z)big)=-pitexti,exp(-k),.$$ Now, note that $$lim_epsilonto0^+,int_Gamma_epsilon,f(z),textdz=pitextitext and lim_epsilonto0^+,int_Gamma_frac1epsilon,f(z),textdz=0,.$$ Because $$I(k)=int_0^infty,fracexp(+textikx)-exp(-textikx)x(x^2+1),textdx-lim_epsilonto0^+,int_Gamma_epsilon,f(z),textdz+lim_epsilonto0^+,int_Gamma_frac1epsilon,f(z),textdz,,$$ we see that $$I(k)=2texti,int_0^infty,fracsin(kx)x(x^2+1),textdx-pitexti,.$$ The result follows immediately.

Approach II.

We apply Richard Feynman's integral trick. First, define $J(k)$ to be the required integral: $$J(k):=int_0^infty,fracsin(kx)x(x^2+1),textdx,.$$ Thus, by the Leibniz Integral Rule, we have $J'(k)=displaystyle int_0^infty,fraccos(kx)x^2+1,textdx$. Let $g(z):=dfracexp(textikz)z^2+1$ for all $zinmathbbCsetminus-texti,+texti$. It follows that $$lim_epsilonto0^+,oint_C_epsilon,g(z),textdz=2,int_0^infty,fraccos(kx)x^2+1,textdx,,$$ where $C_epsilon$ is the positively oriented contour $$left[+epsilon,+frac1epsilonright]cup Biggl,thetain[0,pi]Biggrcupleft[-frac1epsilon,-epsilonright]cup Big,thetain[pi,0]Bigtext for epsilonin(0,1),.$$ Furthermore, we have $$oint_C_epsilon,g(z),textdz=2pitexti,textRes_z=textibig(g(z)big)=pi,exp(-k)text for all epsilonin(0,1)$$ Ergo, $J'(k)=displaystyleint_0^infty,fraccos(kx)x^2+1,textdx=dfracpi2,exp(-k)$. Since $J(0)=0$, $$J(k)=int_0^k,J'(t),textdt=fracpi2,int_0^k,exp(-t),textdt=fracpi2,big(1-exp(-k)big),.$$

Approach III.

It is easy to see that $dfracsin(t)t=displaystylefrac12,int_-1^+1,exp(textittau),textdtau$ for all $tneq 0$. That is, the required integral is given by $$beginalignJ(k):=int_0^infty,fracsin(kx)x,left(x^2+1right),textdx&=frac12,int_0^infty,frackx^2+1,int_-1^+1,exp(textikxt),textdt,textdx\&=frac12,int_-infty^+infty,frackx^2+1,int_0^1,exp(textikxt),textdt,textdx,.endalign$$ Using Fubini's Theorem, we obtain $$J(k)=frac12,int_0^1,k,int_-infty^+infty,fracexp(textikxt)x^2+1,textdx,textdt,.$$
For a real number $R>1$, let $gamma_R$ be the positively oriented contour $$[-R,+R]cupbig,thetain[0,2pi]big,.$$ Then, for $omega geq 0$, we have $$lim_Rtoinfty,oint_gamma_R,fracexp(textiomega z)z^2+1,textdz=int_-infty^+infty,fracexp(textiomega x)x^2+1,textdx=:K(omega),.$$ Ergo, $$K(omega)=2pitexti,textRes_z=textileft(fracexp(textiomega z)z^2+1right)=2pitexti,left(fracexp(-omega)2textiright)=pi,exp(-omega),.$$ As $displaystyle J(k)=frac12,int_0^1,k,K(kt),textdt$, we conclude that $$J(k)=fracpi2,int_0^1,k,exp(-kt),textdt=fracpi2,big(1-exp(-k)big),.$$ From this proof, we also obtain $$int_0^infty,fraccos(kx)x^2+1,textdx=frac12,K(k)=fracpi2,exp(-k),.$$

P.S. I put my solutions in spoilers for people who want to try to solve the problem without being led on by my attempts. If you do not want to waste your time doing what I have already done, then please look at the spoilers. All of the solutions in the spoilers are complex-analytic proofs anyhow, so if you are giving me a real-analytic proof, then there is no way you are going to replicate my work.

HINT:

Let $f(k)$ be given by the integral

$$f(k) =int_0^infty fracsin(kx)x(x^2+1),dx tag1$$

Inasmuch as the improper integral $int_0^infty fracxsin(kx)x^2+1,dx$ converges uniformly for $|k|ge delta>0$, we can differentiate twice under the integral in $(1)$ to reveal

$$f''(k)-f(k)=-fracpi2 textsgn(k)tag2$$

Solve $(2)$ subject to the initial conditions $f(0)=0$ and $f'(0)=fracpi2$.

NOTE:

$$int_0^infty fracxsin(kx)x^2+1,dx=int_0^infty frac(x^2+1-1)sin(kx)x(x^2+1),dx=int_0^infty fracsin(kx)x,dx-int_0^infty fracsin(kx)x(x^2+1),dx$$

You can use a nice property of the Laplace transform to computer your integral:
$$int_0^infty f(x) g(x),dx=int_0^infty mathcalLf(x)(s),mathcalL^-1g(x)(s) ,ds$$
In your case, letting $f(x)=sin (kx)$ and $g(x)=frac1x(x^2+1)$, we can show that
beginalign
int_0^infty fracsin(kx)x(x^2+1),dx &= kint_0^infty frac1-cos(s)k^2+s^2,ds \
&=fracpi2-kint_0^infty fraccos(s)k^2+s^2,ds \
&=fracpi2-frack2int_-infty^infty frace^isk^2+s^2,ds \
&=fracpi2 - frack2 sqrt2pi ,mathcalFleftfrac1k^2+s^2right(omega)Biggr|_omega=1 \
&= fracpi2 - frack2sqrt2pi sqrtfracpi2 frace^-kk \
&=fracpi2-fracpi2e^-k \
&= fracpi2left(1-e^-kright)
endalign
Where we used the Fourier Transform.
Thus,
$$int_0^infty fracsin(kx)x(x^2+1),dx=fracpi2left(1-e^-kright) quad textfor kinmathbbR^+$$