How to compute the Defining integral of cos and power How to calculate the following integral (a , b and H are constants):
$\int_0^{\infty} \cos(ax)\frac{x^{1-2H}}{b^2+x^2}dx$
Thank you!
 A: I think that, without loss of generality, we can assume $a>0$ and $b>0$ if we are in the real domain.$$I=\int_0^{\infty} \cos(ax)\frac{x^{1-2H}}{x^2+b^2}dx$$ does not seem to be an easy integral.
What I tried is firs $x=by$ to make
$$I=b^{-2 H}\int_0^{\infty}\cos (a b y)\frac{y^{1-2 H} }{y^2+1}\,dy$$ then $\alpha=ab$ and $\beta=(1-2H)$ to end with
$$I=b^{-2 H}\int_0^{\infty}\cos (\alpha y)\frac{y^{\beta} }{y^2+1}\,dy$$ which did not help. After a few unsuccessful attempts, I used a CAS which gave for
$$J=\int_0^{\infty}\cos (\alpha y)\frac{y^{\beta} }{y^2+1}\,dy$$
$$J=\frac{  \sec \left(\frac{\pi  \beta }{2}\right)} 2\left( \alpha ^{1-\beta } \sin (\pi  \beta )  \Gamma (\beta -1) \, _1F_2\left(1;\frac{2-\beta
   }{2},\frac{3-\beta}{2};\frac{\alpha ^2}{4}\right)+ \pi 
   \cosh (\alpha )\right)$$ provided that $-1<\beta <2$ that is to say $-\frac 12 < H < 1$.
A: Let $a,b\in \mathbb{R}$, with the condition that $b\neq0$, then, let's consider the similar integral $I(a)$, which will lead us to the desired integral.
\begin{equation}
I(a)=\int\limits_{0}^{+\infty} \frac{\cos(ax)}{x^{2}+b^{2}} \,dx 
\end{equation}
Now, let's differentiate the last expression with respect to $a$ using the Leibniz rule. Then:
\begin{equation}
I'(a)=-\int\limits_{0}^{+\infty} \frac{x\sin(ax)}{x^{2}+b^{2}} \,dx = -\int\limits_{0}^{+\infty} \frac{x^{2}\sin(ax)}{x(x^{2}+b^{2})} \,dx =-\int\limits_{0}^{+\infty} \frac{(x^{2}+b^{2}-b^{2})\sin(ax)}{x(x^{2}+b^{2})} \,dx
\end{equation}
After distributing, we now have two different integrals:
\begin{equation}
I'(t)=-\underbrace{\int\limits_{0}^{+\infty} \frac{\sin(ax)}{x} \,dx}_{P} + b^{2}\underbrace{\int\limits_{0}^{+\infty} \frac{\sin(ax)}{x(x^{2}+b^{2})} \,dx}_{Q}
\end{equation}
Let's deal with $P$ first. Consider the following substitution: $u=ax$, with implies that $du/a=dx$. Also, the limits of integration remain the same. Then:
\begin{equation}
P=-\int\limits_{0}^{+\infty} \frac{\sin(u)}{u} \,du
\end{equation}
This integral is quite known and it has been solved here on various occasions. It converges to $\frac{\pi}{2}$, then $P=-\frac{\pi}{2}$.
We now know that:
\begin{equation}
I'(a)=-\frac{\pi}{2} + b^{2}\int\limits_{0}^{+\infty} \frac{\sin(ax)}{x(x^{2}+b^{2})} \,dx
\end{equation}
Let's differentiate $I(a)$ with respect to $a$ once again, which yields:
\begin{equation}
I''(a)= b^{2}\int\limits_{0}^{+\infty} \frac{x\cos(ax)}{x(x^{2}+b^{2})} \,dx = b^{2}\underbrace{\int\limits_{0}^{+\infty} \frac{\cos(ax)}{(x^{2}+b^{2})} \,dx}_{I(a)}
\end{equation}
Then, in order to find the value of $I(a)$, we need to solve the following differential equation: $I''(a)=b^{2}I(a)$, whose general solution is given by: $I(a)=c_{1}e^{ab}+c_{2}e^{-ab}$. To find the constants, consider the cases $I(a=0)$ and $I'(a=0)$. After solving the system of equation, one finds that $c_{1}=\frac{\pi}{4}\left(\frac{1}{b}+1\right)$ and that $c_{2}=\frac{\pi}{4}\left(\frac{1}{b}+1\right) + \frac{\pi}{2}$.
Then, we finally conclude that:
\begin{equation}
I(a)=\int\limits_{0}^{+\infty} \frac{\cos(ax)}{x^{2}+b^{2}} \,dx= \left(\frac{\pi}{4b}+\frac{\pi}{4}\right)e^{ab} + \left(\frac{\pi}{4b}+\frac{3\pi}{4}\right)e^{-ab}
\end{equation}
If we differentiate $I(a)$ $n$ times, we will obtain the result for the desired integral:
\begin{equation}
I^{(n)}(a)=\int\limits_{0}^{+\infty} \frac{\partial^{n}}{\partial a^{n}}\left[\frac{\cos(ax)}{x^{2}+b^{2}}\right] \,dx= \frac{\partial^{n}}{\partial a^{n}}\left[\left(\frac{\pi}{4b}+\frac{\pi}{4}\right)e^{ab} + \left(\frac{\pi}{4b}+\frac{3\pi}{4}\right)e^{-ab}\right]
\end{equation}
\begin{equation}
I^{(n)}(a)=\int\limits_{0}^{+\infty} (-1)^{n}\frac{\cos(ax)}{x^{2}+b^{2}}x^{n} \,dx= b^{n}\left(\frac{\pi}{4b}+\frac{\pi}{4}\right)e^{ab} + (-1)^{n}b^{n}\left(\frac{\pi}{4b}+\frac{3\pi}{4}\right)e^{-ab}
\end{equation}
\begin{equation}
(-1)^{n}\int\limits_{0}^{+\infty} \frac{\cos(ax)}{x^{2}+b^{2}}x^{n} \,dx=b^{n}\left(\frac{\pi}{4b}+\frac{\pi}{4}\right)e^{ab} + (-1)^{n}b^{n}\left(\frac{\pi}{4b}+\frac{3\pi}{4}\right)e^{-ab}
\end{equation}
\begin{equation}
\int\limits_{0}^{+\infty} \frac{\cos(ax)}{x^{2}+b^{2}}x^{n} \,dx = (-1)^{-n}b^{n}\left(\frac{\pi}{4b}+\frac{\pi}{4}\right)e^{ab} + b^{n}\left(\frac{\pi}{4b}+\frac{3\pi}{4}\right)e^{-ab} \hspace{.5cm} \text{if}\,\, n\in\mathbb{N}_{0}
\end{equation}
If we let $n=1-2H$, then we obtain the desired result:
\begin{equation}
\boxed{\int\limits_{0}^{+\infty} \frac{\cos(ax)}{x^{2}+b^{2}}x^{1-2H} \,dx= (-1)^{2H-1}b^{1-2H}\left(\frac{\pi}{4b}+\frac{\pi}{4}\right)e^{ab} + b^{1-2H}\left(\frac{\pi}{4b}+\frac{3\pi}{4}\right)e^{-ab}}
\end{equation}
