Dtyjlui

Let $x>0$. I have to prove that

$$
int_{0}^{infty}frac{cos x}{x^p}dx=frac{pi}{2Gamma(p)cos(pfrac{pi}{2})}tag{1}
$$

by converting the integral on the left side to a double integral using the expression below:

$$
frac{1}{x^p}=frac{1}{Gamma(p)}int_{0}^{infty}e^{-xt}t^{p-1}dttag{2}
$$

By plugging $(2)$ into $(1)$ I get the following double integral:

$$
frac{1}{Gamma(p)}int_{0}^{infty}int_{0}^{infty}e^{-xt}t^{p-1}cos xdtdxtag{3}
$$

However, I unable to proceed any further as I am unclear as to what method should I use in order to compute this integral. I thought that an appropriate change of variables could transform it into a product of two gamma functions but I cannot see how that would work. Any help would be greatly appreciated.

The Laplace transform of $cos x$ is $frac{s}{1+s^2}$ and the inverse Laplace transform of $frac{1}{x^p}$ is $frac{s^{p-1}}{Gamma(p)}$, hence
$$ int_{0}^{+infty}frac{cos x}{x^p},dx = frac{1}{Gamma(p)}int_{0}^{+infty}frac{s^p}{s^2+1},ds=frac{1}{Gamma(p)}int_{0}^{pi/2}left(tan uright)^p,du $$
equals
$$ begin{eqnarray*}frac{1}{Gamma(p)}int_{0}^{1} v^p (1-v^2)^{-(p+1)/2},dv&=&frac{1}{2,Gamma(p)}int_{0}^{1}w^{(p-1)/2}(1-w)^{-(p+1)/2},dw\& =& frac{Bleft(tfrac{1+p}{2},tfrac{1-p}{2}right)}{2,Gamma(p)}end{eqnarray*} $$
or
$$ frac{Gammaleft(frac{1+p}{2}right)Gammaleft(frac{1-p}{2}right)}{2,Gamma(p)}= frac{pi}{2,Gamma(p)sinleft(frac{pi}{2}(p+1)right)}=frac{pi}{2,Gamma(p)cosleft(frac{pi p}{2}right)}$$
as wanted. We have exploited the Beta function and the reflection formula for the $Gamma$ function.

Your given integral is closely related to the Mellin transform and can be evaluated by using Ramanujan's Master Theorem.

Ramanujan's Master Theorem

Let $f(x)$ be an analytic function with a MacLaurin Expansion of the form
$$f(x)=sum_{k=0}^{infty}frac{phi(k)}{k!}(-x)^k$$then the Mellin Transform of this function is given by
$$int_0^{infty}x^{s-1}f(x)dx=Gamma(s)phi(-s)$$

Therefore expand the cosine function as Taylor series expansion to get

$$begin{align}
mathfrak{I}=int_0^{infty}cos(x)x^{-p}dx&=int_0^{infty}x^{-p}sum_{n=0}^{infty}(-1)^nfrac{x^{2n}}{(2n)!}dx
end{align}$$

In order to bring the above integral in the wanted form for the usage of Ramanujan's Master Theorem apply the substitution $x^2=u$. So we further get

$$begin{align}
mathfrak{I}=int_0^{infty}x^{-p}sum_{n=0}^{infty}(-1)^nfrac{x^{2n}}{(2n)!}dx&=int_0^{infty}x^{-p}sum_{n=0}^{infty}frac{1}{(2n)!}(-x^2)^ndx\
&=int_0^{infty}u^{-p/2}sum_{n=0}^{infty}frac{1}{(2n)!}(-u)^nfrac{du}{2sqrt{u}}\
&=frac12int_0^{infty}u^{-(p+1)/2}sum_{n=0}^{infty}frac{1}{(2n)!}(-u)^ndu\
&=frac12int_0^{infty}u^{-(p+1)/2}sum_{n=0}^{infty}frac{n!/(2n)!}{n!}(-u)^ndu
end{align}$$

By using the relation $Gamma(n)=(n-1)!$ which is valid for all $ninmathbb N$ we can consider the last integral as an application of Ramanujan's Master Theorem with $s=-frac{p-1}2$ and $phi(n)=frac{Gamma(n+1)}{Gamma(2n+1)}$. By finally using the Theorem we obtain

$$begin{align}
mathfrak{I}=frac12int_0^{infty}u^{-(p+1)/2}sum_{n=0}^{infty}frac{n!/(2n)!}{n!}(-u)^ndu&=frac12Gammaleft(-frac{p-1}2right)frac{Gammaleft(frac{p-1}2+1right)}{Gammaleft(2left(frac{p-1}2right)+1right)}\
&=frac1{2Gamma(p)}Gammaleft(1+frac{p-1}2right)Gammaleft(-frac{p-1}2right)
end{align}$$

Now by applying Euler's Reflection Formula with $z=1+frac{p-1}2$ we moreover get

$$begin{align}
mathfrak{I}=frac1{2Gamma(p)}Gammaleft(1+frac{p-1}2right)Gammaleft(-frac{p-1}2right)&=frac1{2Gamma(p)}frac{pi}{sinleft(pileft(1+frac{p-1}2right)right)}\
&=frac1{2Gamma(p)}frac{pi}{sinleft(frac{ppi}2+frac{pi}2right)}\
&=frac1{2Gamma(p)}frac{pi}{cosleft(frac{ppi}2right)}
end{align}$$

where within the last step the fundamental relation $sinleft(x+frac{pi}2right)=cos(x)$ was used. Thus for the original integral $mathfrak{I}$ we get

$$mathfrak{I}=int_0^{infty}cos(x)x^{-p}dx=frac{pi}{2Gamma(p)cosleft(pfrac{pi}2right)}$$

Hint: $displaystyleint_{0}^{infty}frac{cos x}{x^p}dx= text{Real part of}:int_{0}^{infty}frac{e^{iz}}{z^p}dz$ and use residue theorem. This has a pole of order $p$ hence the term $Gamma (p)$ in the denominator.

So let us follow your initial line of thought and convert the integral to a double integral. As you correctly observe, as
$$frac{1}{x^p} = frac{1}{Gamma (p)} int_0^infty e^{-xt} t^{p - 1} , dt,$$
which, by the way, is just the Laplace transform for the function $x^{p -1}$, as a double integral your integral can be rewritten as
$$int_0^infty frac{cos x}{x^p} , dx = frac{1}{Gamma (p)} int_0^infty int_0^infty e^{-xt} cos x t^{p - 1} , dt , dx,$$
or
$$int_0^infty frac{cos x}{x^p} , dx = frac{1}{Gamma (p)} int_0^infty t^{p - 1} int_0^infty e^{-xt} cos x , dx , dt,$$
after changing the order of integration.

The inner $x$-integral can be readily found. Either using integration by parts twice, or recognising the integral as the Laplace transform for the function $cos x$, as
$$int_0^infty e^{-xt} cos x , dx = frac{t}{1 + t^2},$$
we have
$$int_0^infty frac{cos x}{x^p} , dx = frac{1}{Gamma (p)} int_0^infty frac{t^p}{1 + t^2} , dt.$$
Enforcing a substitution of $t mapsto sqrt{t}$ leads to
begin{align}
int_0^infty frac{cos x}{x^p} , dx &= frac{1}{2 Gamma (p)} int_0^infty frac{t^{frac{p}{2} - frac{1}{2}}}{1 + t} , dt = frac{1}{2 Gamma (p)} int_0^infty frac{t^{frac{p + 1}{2} - 1}}{(1 + t)^{frac{p + 1}{2} + frac{1 - p}{2}}}.
end{align}
As this is exactly of the form of the Beta function (see the second of the integral representations in the link) we have
begin{align}
int_0^infty frac{cos x}{x^p} , dx &= frac{1}{2 Gamma (p)} text{B} left (frac{p + 1}{2}, frac{1 - p}{2} right )\
&= frac{1}{2 Gamma (p)} Gamma left (frac{p}{2} + frac{1}{2} right ) Gamma left (frac{1}{2} - frac{p}{2} right ) \
&= frac{1}{2 Gamma (p)} Gamma left (frac{p}{2} + frac{1}{2} right ) Gamma left [1 - left (frac{p}{2} + frac{1}{2} right ) right ] \
&= frac{1}{2 Gamma (p)} frac{pi}{sin (p + 1)pi/2} qquad (*)\
&= frac{pi}{2 Gamma (p) cos left (frac{pi p}{2} right )},
end{align}
as required. Note Euler's reflection formula was used in ($*$).