Compute $\int_0^\infty \frac{\sin(ax)}{b^2+x^2}~dx$, where $a>0, b>0$

Question

If the $\sin(ax)$ function is replaced by $\cos (ax)$, it's well known and can be solved by many ways. But if we want to solve the sine version integral, i.e.

$$I=\int_0^\infty \frac{\sin(ax)}{b^2+x^2}~dx$$

It seems not that trivial as thought. Let $x=bt, k=ab,$

$$I=\frac{1}b\int_0^\infty \frac{\sin(kt)}{1+t^2}~dt,~~~~F=F(k)=\int_0^\infty \frac{\sin(kt)}{1+t^2}~dt$$ Now, we compute $F$:

$$F'=\int_0^\infty \frac{t\cos(kt)}{1+t^2}~dt~~ \overset{\theta=kt}{\longrightarrow} ~~F'=\int_0^\infty \frac{\theta\cos(\theta)}{k^2+\theta^2}d\theta$$

Take second derivative

$$F''=-\int_0^\infty \frac{2k\theta\cos(\theta)}{(k^2+\theta^2)^2}d\theta=\int_0^\infty k\cos(\theta)d\left(\frac{1}{k^2+\theta^2}\right)$$ Integration by part

$$F''=-\frac{1}k+\int_0^\infty \frac{k\sin(\theta)}{k^2+\theta^2}d\theta~~ \overset{\theta=kt}{\longrightarrow} ~~F''=-\frac{1}k+\int_0^\infty \frac{\sin(kt)}{1+t^2}dt$$ Therefore,

$$F''(k)=-\frac{1}k+F(k)$$

Solve this 2nd order inhomogeneous differential equation and we get

$$F(k)=c_1 e^k+c_2 e^{-k}+\text{P.V}\left(\frac{1}2e^k\int_k^\infty \frac{e^{-t}}{t}dt+\frac{1}2e^{-k}\int^k_{-\infty} \frac{e^{t}}{t}dt\right)$$

Define: $\displaystyle\text{Ei}(z)=\text{P.V}\left(-\int_{-z}^\infty \frac{e^{-t}}{t}dt\right)$, we get

$$F(k)=c_1 e^k+c_2 e^{-k}+\frac{1}2\left(-e^k\text{Ei}(-k)+e^{-k}\text{Ei}(k)\right)$$

From the integral $\displaystyle F(k)=\int_0^\infty \frac{\sin(kt)}{1+t^2}~dt$ we can see $F(0)=0$ and $\displaystyle\lim_{k\to\infty} F(k)$ is bounded, hence, $c_1=c_2=0$

$$\int_0^\infty \frac{\sin(kt)}{1+t^2}~dt=\frac{-e^k\text{Ei}(-k)+e^{-k}\text{Ei}(k)}2$$

Finally,

$$\boxed{\int_0^\infty \frac{\sin(ax)}{b^2+x^2}~dx=\frac{-e^{ab}\text{Ei}(-ab)+e^{-ab}\text{Ei}(ab)}{2b}}$$

It seems this is the simplest form I can get. Are there other simple ways to solve this integral, such as contour integral, etc?

Answer 1

A way to do it is to write $$\frac 1{x^2+b^2}=\frac 1{(x-ib)(x+ib)}=\frac{i}{2b}\left(\frac{1}{x+i b}-\frac{1}{x-i b}\right)$$ that is to say $$\frac 1{x^2+b^2}=i\frac{a}{2b}\left(\frac{1}{ax+i ab}-\frac{1}{ax-i ab}\right)$$

Consider the first one $$\int \frac {\sin(ax)}{ax+iab}\,dx=\int \frac {\sin(ax +iab-iab)}{ax+iab}\,dx=\frac 1 a\int \frac {\sin( t-i ab)} t\,dt$$ Expand the sine $$\sin(t-iab)= \cosh (a b)\,\sin (t)-i \sinh (a b)\,\cos (t) $$ Do the same for the second one and rearrange to obtain by the end $$2b \int \frac {\sin(ax)}{x^2+b^2}\,dx=\sinh (a b) (\text{Ci}(a (x-i b))+\text{Ci}(a (i b+x)))+$$ $$i \cosh (a b) (\text{Si}(a (i b+x))+\text{Si}(a(i b- x)))$$

Use the bounds to finish.

Another solution is to write $$\int \frac {\sin(ax)}{x^2+b^2}\,dx=\Im \left(\int \frac {e^{i ax}}{x^2+b^2}\,dx\right)$$ Using exactly the same process $$2b\int \frac {e^{i ax}}{x^2+b^2}\,dx=i e^{a b} \text{Ei}(i a x-a b)-i e^{-a b} \text{Ei}(a (b+i x))$$ to obtain your result.