How to prove that $ \int_{0}^{\infty} x^n e^{-xt} \sin x\frac{dx}{x} =\frac{i(n-1)!}{2(1+t^2)^n}\left((t-i)^n-(t+i)^n\right)$ Let $n\ge 1$ How to prove that $$I_n(t)=  \int_{0}^{\infty} x^n e^{-xt} \sin x\frac{dx}{x} =\frac{(n-1)!}{(1+t^2)^n}\frac{\left((t+i)^n-(t-i)^n\right)}{2i}$$

I have manage to prove that one can apply Lebesgue theorem for differentiability and I came across the following relation 
  $$I'_n(t)=-I_{n+1}(t)$$ But this seems does not helps me so far using induction. 

Can one helps me form here or Is there possible way to derive directly this integral?
 A: Write $\displaystyle \mathcal{I}(t) = \int_0^{\infty} e^{-xt}\sin{x}\,{dx} = \frac{1}{t^2+1}$. Differentiate both sides $n-1$ times. 
A: Given,
$$I_n(t)=  \int_{0}^{\infty}(x^{n-1}\sin x)e^{-xt}~\mathrm dx$$
This is the Laplace transform of $x^{n-1}\sin x$ with parameter $t$, i.e., $I_n(t)=\mathcal L\{x^{n-1}\sin x~;~t\}$
Now, $$\mathcal L\{\sin x~;~t\}=\frac 1{1+t^2}=(\arctan t)'=L(t)~\textrm{(say)}$$
Now, multiplication by $x^{n-1}$ makes the Laplace transform $$\mathcal L\{x^{n-1}\sin x~;~t\}=(-1)^{n-1}L^{(n-1)}(t)=(-1)^{n-1}(\arctan t)^{(n)}$$
For the closed form of the $n^{\textrm{th}}$ derivative of $\arctan$, see here, which I think looks very similar to what you have here in the original post.
A: Noting
$$ \sin x=\frac{1}{2i}(e^{ix}-e^{-ix}), \int_0^\infty x^{n-1}e^{-ax}dx=a^{-n}(n-1)! $$
one has
\begin{eqnarray}
&& \int_{0}^{\infty} x^n e^{-xt} \sin x\frac{dx}{x}\\
&=& \frac{1}{2i}\int_{0}^{\infty} x^{n-1} e^{-xt}(e^{ix}-e^{-ix})dx\\
&=& \frac{1}{2i}\left[\int_{0}^{\infty} x^{n-1} e^{-x(t-i)}dx-\int_{0}^{\infty} x^{n-1} e^{-x(t+i)}dx\right]\\
&=& \frac{1}{2i}\left[(t-i)^{-n}-(t+i)^{-n}\right](n-1)!\\
&=&\frac{i(n-1)!}{2(1+t^2)^n}\left((t-i)^n-(t+i)^n\right).
\end{eqnarray}
A: Edit. First version lacked clarity and had some errors.
First note that
\begin{align} 
I_n(t) =\frac{1} {i2} \int^\infty_0 x^{n-1}e^{(i-t) x} \,dx - \frac{1} {i2} \int^\infty_0 x^{n-1}e^{(-i-t) x} \,dx
\end{align} 
We have two integrals. Both can be related to a contour integral. One also can see somehow the $\Gamma$-function in it, so it would be very nice to expres these integrals in terms of $\Gamma$-function. That is what we will do and we only show it for the first integral. For that consider:
\begin{align} 
\tag{1}
\oint_C z^{n-1}e^z\, dz
\end{align} 
The contour $C$ is a circle sector with angle $\arctan(1/t)$ going from the origin all the way up in the second quadrant and coming back all the way down through the negative real axis. Let the radius of the "circle" be $R$. The contour can be seen below.
$\hspace{70pt}  $
When $R$ goes to infinity the circular contribution goes to zero by the ML-lemma. Since the integrand is holomorphic, by Cauchy's theorem we get that the integral in $(1)$ is zero and by parameterization (and taking     $R\to \infty $) :
\begin{align} 
\int^0_{-\infty}x^{n-1}e^x\, dx+(i-t)^n\int^\infty_0 x^{n-1}e^{(i-t) x} \, dx=0
\end{align}
We substitute $u=-x$ for the first integral:
\begin{align}
\int^\infty_{0}(-1)^{n-1}u^{n-1} e^{-u}\,du + (i-t)^n\int^\infty_0 x^{n-1}e^{(i-t) x} \, dx=0
\end{align} 
Ladies and gentleman, that is what we just wanted: the first integral is the $\Gamma$-function and all that implies:
\begin{align} 
\frac{1} {i2} \int^\infty_0 x^{n-1}e^{(i-t) x} \,dx=\frac{(n-1)!(-1)^{n}} {i2(i-t)^n} =\frac{(n-1)!(t+i)^n}{i2(1+t^2)^n}
\end{align} 
Similary we get:
\begin{align} 
\frac{1} {i2} \int^\infty_0 x^{n-1}e^{(-i-t) x} \,dx=\frac{(n-1)!(t-i)^n}{i2(1+t^2)^n}
\end{align} 
Adding them up yields the desired result.  
