Mellin Transform of Sine and Cosine

            Today I want to compute two pairs of integrals using two different techniques. Firts pair is the Mellin transform of sine and cosine using contour integration. The second one is a generalization of the first pair and will be evaluated through a mix of real and complex methods. In the end we will see how we can recover the first result from the second.

The Mellin transforms of  \cos(kx) and  \sin(kx) are given by the following integrals:

\int_{0}^{\infty}\cos(kx)x^{a-1}=\frac{\Gamma(a)}{k^a}\cos\left( \frac{ a\pi}{2}\right)

\int_{0}^{\infty}\sin(kx)x^{a-1}=\frac{\Gamma(a)}{k^a}\sin\left( \frac{ a\pi}{2}\right)

We first consider the following integral and the equate Real and Imaginary parts to get the desired results

I=\int_{0}^{\infty}\frac{e^{ikx}}{x^a}dx

for 0<a<2

Consider the following integral in the complex plain.

\oint_C\frac{e^{ikz}}{z^a}dz

For k>0. C is the contour below


Since there is no singularity inside the contour, by Cauchy´s theorem

\oint_C\frac{e^{ikz}}{z^a}dz=0

\oint_C\frac{e^{ikz}}{z^a}dz=\int_{\epsilon}^{R}\frac{e^{ikx}}{x^a}dx+\int_{C_{R}}\frac{e^{ikz}}{z^a}dz+\int_{i R}^{i \epsilon}\frac{e^{iky}}{y^a}dy+\int_{C_{\epsilon}}\frac{e^{ikz}}{z^a}dz=0

For the integral around the big arc let z=Re^{i \theta}}we have

\int_{C_{R}}\frac{e^{ikz}}{z^a}dz=\int_{0}^{\pi/2}\frac{e^{ikRe^{i \theta}}}{R^ae^{ia \theta}}iRe^{i \theta}d \theta

= R^{1-a}\int_{0}^{\pi/2}e^{-kR \sin (\theta)}e^{i \cos( \theta)}e^{i \theta (1-a)}d \theta \leq R^{1-a}\int_{0}^{\pi/2}e^{-kR \sin (\theta)}d \theta

Since k>0 as R \rightarrow \infty this integral vanishes

Similarly, letting z=\epsilon e^{i \phi} in thefourth integral

\int_{C_{\epsilon}}\frac{e^{ikz}}{z^a}dz=-\epsilon^{1-a} \int_{0}^{\pi/2}e^{ik \epsilon e^{i \phi}}e^{i \phi (1-a)} d \phi < -\epsilon^{1-a} \int_{0}^{\pi/2}e^{-k \epsilon \sin( \phi) } d \phi

The integral is finite and as \epsilon \rightarrow 0 this term vanishes

Therefore, after taking the limits R \rightarrow \infty and \epsilon \rightarrow 0 we have

\int_{0}^{\infty}\frac{e^{ikx}}{x^a}dx=\int_{0}^{i \infty}\frac{e^{iky}}{y^a}dy

Now let iky=-t in the integral on the right hand side.

-y = \frac{t}{ik} \Rightarrow dy=-\frac{dt}{ik}

\text{ when} \,\,\, y=0 \Rightarrow t=0

\text{ when} \,\,\, y=i \infty \Rightarrow t= \infty

Therefore, we get

\int_{0}^{\infty}\frac{e^{ikx}}{x^a}dx=(-ik)^{a-1}\int_{0}^{ \infty}\frac{e^{-t}}{t^a}dt=(-ik)^{a-1} \Gamma\left( 1- a\right)

\int_{0}^{\infty}\frac{\cos(kx)}{x^a}dx+\int_{0}^{\infty}\frac{\sin(kx)}{x^a}dx=k^{a-1}e^{-\frac{i \pi}{2}(a-1)} \Gamma\left( 1- a\right)=k^{a-1}\Gamma\left( 1- a\right)\left(\cos\left(\frac{ \pi}{2}(a-1)}\right)-i\sin\left(\frac{ \pi}{2}}(a-1)\right) \right)

Equating Real and Imaginary parts

\int_{0}^{\infty}\frac{\cos(kx)}{x^a}dx=k^{a-1}\cos\left(\frac{ \pi}{2}(a-1)}\right) \right)\Gamma\left( 1- a\right)=k^{a-1}\cos\left(\frac{ \pi}{2}(1-a)}\right) \right)\Gamma\left( 1- a\right)(1)

\int_{0}^{\infty}\frac{\sin(kx)}{x^a}dx=-k^{a-1}\sin\left(\frac{ \pi}{2}}(a-1)\right) \right)\Gamma\left( 1- a\right)=k^{a-1}\sin\left(\frac{ \pi}{2}}(1-a)\right) \right)\Gamma\left( 1- a\right)(2)

Since \cos(-x)=\cos(x) and sin(-x)=-\sin(x)

Letting a\rightarrow 1-a we get

\int_{0}^{\infty}\cos(kx)x^{a-1}dx=\frac{\Gamma(a)}{k^a}\cos\left( \frac{ a\pi}{2}\right)(3)

\int_{0}^{\infty}\sin(kx)x^{a-1}dx=\frac{\Gamma(a)}{k^a}\sin\left( \frac{ a\pi}{2}\right)(4)

We now evaluate another very interesting and useful pair of integrals which allow us to recover the same result above

Consider the pair of integrals

\int_{0}^{\infty} e^{-a t} t^{x-1} \cos (b t) d t=\Gamma(x) \frac{\cos \left(x \arctan \left(\frac{b}{a}\right)\right)}{\left(a^{2}+b^{2}\right)^{\frac{x}{2}}}

\int_{0}^{\infty} e^{-a t} t^{x-1} \sin (b t) d t=\Gamma(x) \frac{\sin \left(x \arctan \left(\frac{b}{a}\right)\right)}{\left(a^{2}+b^{2}\right)^{\frac{x}{2}}}

To proof the pair of integrals recall that for a>0 and b>0

a+i b=r e^{i \theta}=\left(a^{2}+b^{2}\right)^{1 / 2} e^{i \tan ^{-1}\left(\frac{b}{a}\right)}(5)

and

\int_{0}^{\infty} e^{-s t} t^{x-1} d t=\frac{\Gamma(x)}{s^{x}}(6)

(6) in obtained by letting s t=u Consider the following integral

I=\int_{0}^{\infty} e^{-a t} t^{x-1} e^{-i b t} d t=\int_{0}^{\infty} e^{-a t} t^{x-1}(\cos (b t)-i \sin (b t)) d t

I=\int_{0}^{\infty} e^{-a t} t^{x-1} e^{-i b t} d t=\int_{0}^{\infty} e^{-(a+i b) t} t^{x-1} dt

letting a+i b=s and using (6)

I=\int_{0}^{\infty} e^{-s t} t^{x-1} d t=\frac{\Gamma(x)}{s^{x}}=\frac{\Gamma(x)}{(a+i b)^{x}}(7)

Now from (5) we have

(a+i b)^{x}=\left(\left(a^{2}+b^{2}\right)^{1 / 2} e^{i \tan ^{-1}\left(\frac{6}{a}\right)}\right)^{x}(8)

Plugging (8) in (7) we get

\int_{0}^{\infty} e^{-a t} t^{x-1} e^{-i b t} d t=\frac{\Gamma(x) e^{-i x \tan ^{-1}\left(\frac{b}{a}\right)}}{\left(a^{2}+b^{2}\right)^{x / 2}}=\frac{\Gamma(x)\left(\cos \left(x \arctan \left(\frac{b}{a}\right)\right)-i \sin \left(x \arctan \left(\frac{b}{a}\right)\right)\right)}{\left(a^{2}+b^{2}\right)^{x / 2}}

Equating Real and Imaginary part

\int_{0}^{\infty} e^{-a t} t^{x-1} \cos (b t) d t=\Gamma(x) \frac{\cos \left(x \arctan \left(\frac{b}{a}\right)\right)}{\left(a^{2}+b^{2}\right)^{\frac{x}{2}}}(9)

\int_{0}^{\infty} e^{-a t} t^{x-1} \sin (b t) d t=\Gamma(x) \frac{\sin \left(x \arctan \left(\frac{b}{a}\right)\right)}{\left(a^{2}+b^{2}\right)^{\frac{x}{2}}}(10)

If we let a=0 in (9) and (10) and

\lim_{a \rightarrow 0}\arctan \left(\frac{b}{a}\right)\rightarrow \frac{\pi}{2}

We recover (3) and (4)

\int_{0}^{\infty}\cos(bt)t^{x-1}dt=\frac{\Gamma(x)}{b^x}\cos\left( \frac{ x\pi}{2}\right)

\int_{0}^{\infty}\sin(bt)t^{x-1}dt=\frac{\Gamma(x)}{x^b}\sin\left( \frac{ x\pi}{2}\right)

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