# How would you evaluate $I:=\int_ {0}^{\infty} \frac {\cos(ax)} {(x^2 + b^2)^n} \ \mathrm{d}x$?

Any pointers on how should I start?

$$I:=\int_ {0}^{\infty} \frac {\cos(ax)} {(x^2 + b^2)^n} \ \mathrm{d}x$$

#### Solutions Collecting From Web of "How would you evaluate $I:=\int_ {0}^{\infty} \frac {\cos(ax)} {(x^2 + b^2)^n} \ \mathrm{d}x$?"

I will leave some details to the reader. W.L.o.G. assume $a,b>0$. Consider
$$J(a,b) = \int_0^{\infty} \frac{\cos(ax)}{x^2+b^2} \ \mathrm{d}x,$$
then
$$\frac{\partial^{n-1} J}{\partial b^{n-1}} = (-2b)^{n-1}(n-1)!\int_0^{\infty} \frac{\cos(ax)}{(x^2+b^2)^n} \ \mathrm{d}x.$$
The integral $J$ can be calculated via the Residue Theorem as
$$J = \frac{\pi}{2b}\mathrm{e}^{-ab},$$
so we have
$$\int_0^{\infty} \frac{\cos(ax)}{(x^2+b^2)^n} \ \mathrm{d}x = \frac{(-2b)^{1-n}}{(n-1)!}\frac{\partial^{n-1} }{\partial b^{n-1}}\left(\frac{\pi}{2b}\mathrm{e}^{-ab}\right),$$
where
$$\frac{\partial^{n-1} }{\partial b^{n-1}}\left(\frac{\mathrm{e}^{-ab}}{b}\right) = (-1)^{n-1}\frac{\mathrm{e}^{-ab}}{b^{n}}\displaystyle \sum_{i=0}^{n-1}\frac{(n-1)!}{(n-i-1)!} (ab)^{n-i-1},$$
so the final result is, with some cancellation –
$$\int_0^{\infty} \frac{\cos(ax)}{(x^2+b^2)^n} \ \mathrm{d}x = \frac{\pi\mathrm{e}^{-ab}}{2^{n}b^{2n-1}}\displaystyle \sum_{k=1}^{n}\frac{(ab)^{n-k}}{(n-k)!} , \qquad n \ge 1.$$

One may also do this one using residues, though the resulting series can be rather tedious.

But, it follows a binomial pattern.

I am going to allow the power on the denominator to be $n+1$ instead of $n$ as to find the nth derivative.

Consider $$\int_{C}\frac{e^{iaz}}{(z^{2}+b^{2})^{n+1}}$$, where C is the semicircular contour in the UHP with poles at $\pm bi$, which have order $n+1$.

The only pole to be concerned with is the one at $bi$.

Thus, by taking the $nth$ derivative and defining $I(z)=\frac{e^{iaz}}{(z+bi)^{n+1}}$.

$$I^{(n)}(z)=\frac{(ia)^{n}e^{iaz}}{(z+bi)^{n+1}}-\frac{(n+1)(n)(ia)^{n-1}e^{iaz}}{(z+bi)^{n+2}}+\frac{(n)(n-1)(ia)^{n-2}e^{iaz}(n+1)(n+2)}{2!(z+bi)^{n+3}}-\cdot\cdot\cdot +\frac{(-1)^{n}e^{iaz}(n+1)(n+2)\cdot\cdot\cdot (2n)}{(z+bi)^{2n+1}}$$.

Since $$\int\frac{e^{iaz}}{(z^{2}+b^{2})^{n+1}}dz=\frac{2\pi i}{n!}I^{(n)}(bi)$$,

let $z\to bi$, simplify a little, and note the powers of the $i$ terms eliminate the alternation of the series.

Several terms being:

$$\frac{2\pi i}{n!}\left[\frac{e^{-ab}}{(2bi)^{n+1}}(ia)^{n}-\frac{ne^{-ab}(n+1)}{(2bi)^{n+2}}(ia)^{n-1}+\cdot\cdot\cdot\right]$$

simplify a little:

$$\int\frac{e^{iaz}}{(z^{2}+b^{2})^{n+1}}dz=\frac{2\pi e^{-ab}}{n!}\left[\frac{a^{n}}{(2b)^{n+1}}+\frac{n(n+1)}{(2b)^{n+2}}a^{n-1}+\frac{(n-1)n(n+1)(n+2)}{2!(2b)^{n+3}}a^{n-2}+\cdot\cdot\cdot +\frac{(2n)!}{n!(2b)^{2n+1}}\right]$$

The portions along the x axis and around the big arc:

$$\int_{-\infty}^{0}\frac{e^{iax}}{(b^{2}+x^{2})^{n+1}}dx+\int_{0}^{\infty}\frac{e^{iax}} {(b^{2}+x^{2})^{n+1}}dx+\int_{0}^{\pi}\frac{e^{ia Re^{i\theta}}}{(b^{2}+R^{2}e^{2i\theta})^{n+1}}iRe^{i\theta}d\theta$$.

By letting $x\to -x$ in the first one and adding to the second, one obtains:

$$2\int_{0}^{\infty}\frac{\cos(ax)}{(b^{2}+x^{2})^{n+1}}dx$$

Of course, the arc tends to 0 due to the standard $e^{-aR\sin\theta}\to 0$, as $R\to \infty$ argument.
Thus, putting it together, the result becomes:

$$\int_{0}^{\infty}\frac{\cos(ax)}{(b^{2}+x^{2})^{n+1}}dx=\frac{\pi e^{-ab}}{n!(2b)^{2n+1}}\left[(2ab)^{n}+n(n+1)(2an)^{n-1}+\frac{(n-1)n(n+1)(n+2)}{2!}(2ab)^{n-2}+\cdot\cdot\cdot \frac{(2n)!}{n!}\right]$$

which can be written in general as:

$$\int_{0}^{\infty}\frac{\cos(ax)}{(x^{2}+b^{2})^{n+1}}dx=\frac{\pi e^{-ab}}{n! (2b)^{2n+1}}\sum_{k=0}^{n}\frac{(2ab)^{n-k}(n+k)!}{k!(n-k)!}$$