Math Examples of Fourier Series and Laplace Series

For Mellin Transform value we are going to follow the below steps:
Step 1: In the first step we use Mellin transform definition for given value,
First we write the given requirement about mellin value,
^(s) = _n=1∑^+∞ λ_kµ_k^-s,
Here we have given polylogarithm function
 ^(s) = _n=1∑^+∞ d(k)/k^-s ξ²(s),
Where ‘w’ and ‘k’ are equal values which belong to natural Numbers.
And here λ_k =d(k), µ_k = k   and g(x) = e^-x all values are use in Mellin definition .
 D*(s) = Г(s)ξ²(s) , where s is belongs to (1,+∞),
Step 2: In the second step we are going to explain the Mellin series.
Now we,
 D*(s)^[_n=0∑^+∞ 1/(s-1)² + y/(s-1)] + [1/4s]-_n=0∑^+∞ ξ² (-2k - 1)/(2k +1)! 1/s+ 2k +1 where s is belongs to (-∞ to +∞),
Step 3: Now we separate the resulting function in Mellin form.
D(x) 1/x (-log x + y) + ¼ - _k=0∑^+∞ ξ² ((-2k -1)/(2k +1)!)x^2k+1,
At last we get Mellin transform value.

For Mellin Transform value we are going to follow the below steps:
Step 1: In the first step we use Mellin transform definition for given value.
First we write the given requirement about mellin value,
L_w(x) = _n=1∑^+∞ e^-nx/n^w,
Here we have given poly Logarithm function,
 Li_w (z) = _n=1∑^+∞ zⁿ n^-w,
Where ‘w’ and ‘k’ is equal value which are belongs to natural number.
And here λ_n = 1/n^w  , µ_k = n   and g(x) = e^-x all values are use in Mellin definition .
^(s) = ξ(s + k), L_k*(s) = ξ(s + k)Г(s) , where s is belongs to (1, +∞),
Step 2: In the second step we are going to explain the Mellin series.
Now we,
 L_k*(s)^ _n=0∑^+∞ (-1)ⁿ ξ(k-n)/n! 1/(s+n) + (-1)^k-1/(k-1)![1/(s+k -1)² + H_k-1/s+k-1] where s is belongs to (-∞ to +∞)
Step 3: Now we separate the resulting function
L_k(x) = (-1)^k-1/(k-1)! x^k-1 [-log x + H_k-1]+ _n=0∑^+∞ (-1)ⁿ ξ(k-n)/n! xⁿ,
L_1/2 (x) = √(л/x) + _n=0∑^+∞ (-1)ⁿ ξ(1/2 - n)/n! xⁿ .
(Here we take k= ½ then after we are resulting the value at k=1/2) by these expression we got L_1/2 value.
At last we get Mellin transform value.

Here we are going to follow the below steps.
Step 1: In the first step we use to given function in mellin form.
l(x) = log Г(x+1) – λx = _n=1∑^+∞ [x/n – log (1+ x/n)] where s is belongs to -2 to +∞,  
Here by this expression we got the  λ_n = 1 , µ_n = 1/n ,
 g(x)= x-log(1+x),
Step 2: In this step we write the expression in Mellin Transform from.
^(s) = _n=1∑^+∞ λ_k µ_k^-s = _n=1∑^+∞ n^s = ξ(-s),
l*(s) = ^(s)g*(s) = -ξ(-s) л/s sin лs,
In this way we get Mellin transform value.

For solving the Mellin Transform function we are going to follow the below steps:
Step 1: In the first step we explain function by using given value.
^(s) = _k=1∑^+∞ λ_k µ_k^-s = _k=1∑^+∞ k ^-1+s  =ξ(1-s),
Step 2: In this step we are going to use the Mellin transform properties.
H*(s) = ^(s) g*(s) = -ξ(1-s) л/sin лs  s is belong to -1 to 0  [after using Mellin transform we got mellin transfrom].
ξ(s) = 1/s-1 +λ +….., ξ(1-s) = -1/s +λ (Here we got ξ(s) value by mellin expression),
h*(s)^ [1/s² – λ/s] – _k=1∑^+∞ (-1)^k ξ(1-k)/(s-k)   where s is belongs to -1 to +∞,
H_n  = log n + у + _k≥1∑ (-1)^k B_k / k 1/(n^k),
= log n +  у + 1/2n -1/2n +1/120 n⁴ ……,
After solving we got H_n series in Mellin transform series.

For solving Mellin Transform we need to follow the below steps:
Step 1: In the first step we explain given Exponential Function
f(x) = e^-x = 1- x+ x²/2! –x³/3! + …….[After explanation exponential function],
This expression can be write like = _x→0∑^m_j=0 (-1)^j / j! x^j + o(x^m+1),
Step 2: In this step we used the Mellin transform definition,
f*(s) = Г(s) ,  where x is belong to all Integer limit (0 , +∞),
f*(s) is meromorphically containable to (-M -1 , +∞),
f*(s) ^ _x→0∑^m_j (-1)^j/j! 1/(s+j) where s is belong limit (-M , -1 , +∞)
Step 3: In this step we finalize the resulting expression
Finally we get,
Г(s) ^_j=0∑^∞ (-1)^j /j! x 1/s+j  , where s is belong to all integer value.
In this way we got final value of Mellin transform is  
Г(s) ^_j=0∑^∞ (-1)^j /j! x 1/s+j. 

For solving this problem we need to follow the below steps:
Step 1: In the first step, we write the definition of Mellin Transform function,
M [∑_k λ_k f(µ_kx); s] = (∑_k λ_k µ^-s)f*(s),
Step 2: In this step we find the value of λ_k , µ_k , f(x) = e^-x,
λ_k  =1, µ_k  =k , f(x) = e^-x (here we got the value of λ_k, µ_k by mellin transform properties.
g*(s) = (1/1^s + 1/2^s + 1/3^s ….) [This Mellin transform expression],
M[e^-x ;s]  
Now we get final value of mellin transform value which is g*(s) = (1/1^s + 1/2^s + 1/3^s …….).

Let us suppose that f(x) be a function with Period 2π defined in the interval a<x<a<2 π then the infinite trigonometric series given by,
 f (x) = a₀ + ∑_n=1∞ [a_n cos nx + b_nsin nx], with Fourier coefficient a_0, a_n, b_n is known as Fourier series for the function f(x). The values of  a_0, a_n, b_n are given by the following Euler’s formulae,  
a₀=1/2 π∫_a^a+2π f(x)dx,
a _n = 1/π ∫_a^a+2π f(x)cos(nx)dx,
b_n= 1/π ∫_a^a+2π f(x)sin(nx)dx.
Now the function is f(x)= sin ax, since sin ax is an odd function,
The Fourier series of sin ax =∑_n=1∞ b_nsin nx,
b_n= 2/π ∫₀^π f(x)sin(nx)dx,
b_n= 2/π ∫₀^π sin ax sin(nx)dx,
=1/π ∫₀^π[cos(n-a)x-cos(n+a)x]dx, [ because 2sinA sinB=cos(A-B)-cos(A-B)],
=1/π[sin(n-a)x/n-a – sin(n+a)x/n+a]₀^π,
=1/π[sin(n-a)π/n-a – sin(n+a)π/n+a],
=1/π[(sin nπ cos aπ – cos nπ sin aπ)/(n-a) -(sin nπ cos aπ + cos nπ sin aπ)/(n+a)],
=– (cos nπ sin aπ)/π[1/n-a +1/n+a],
=-(-1)ⁿ/π (2n sin aπ)/n²-a²,
Putting n=1, 2, 3, 4…………………………..
b₁=(2 sin aπ)/π.1/1²-a²,
b₂=-(2 sin aπ)/π.2/2²-a² etc.
sin ax = (2 sin aπ)/π[sinx/1²-a²-2sinx/2²-a²…………………].

Let us suppose that f(x) be a function with Period 2π defined in the interval a<x<a<2 π then the infinite trigonometric series given by,
 f (x) = a₀ + ∑_n=1∞ [a_n cos nx + b_nsin nx], with Fourier coefficient a_0, a_n, b_n is known as Fourier series for the function f(x). The values of a_0, a_n, b_n are given by the following Euler’s formulae,  
a₀=1/2 π∫_a^a+2π f(x)dx,
a _n = 1/π ∫_a^a+2π f(x)cos(nx)dx,
b_n= 1/π ∫_a^a+2π f(x)sin(nx)dx.
Now the function is f(x) = π+ x;-π<x<0, π- x;-0<x<π,
So,
a₀=1/2 π∫_-π^π f(x)dx,
=1/2 π[∫_-π⁰ f(x)dx+1/2 π∫₀^π f(x)dx],
=1/2 π[∫_-π⁰­­(π+ x) dx+1/2 π∫₀^π(π- x) dx],
=1/2 π[πx+x²/2]_-π⁰+1/2 π[πx-x²/2]₀^π,
a₀=1/2 π[π-x²/2+π-x²/2]_,
a₀= π/2.
a _n =  1/π ∫_-π^π f(x)cos(nx)dx,
=1/2 π[∫_-π⁰­­(π+ x) cosnx dx+1/2 π∫₀^π(π- x) cosnx  dx],
=1/π[(π+x)sin nx/n + cosnx/n²]_-π⁰+1/2 π[(π-x) sin nx/n - cosnx/n²]₀^π, [integration by parts]
=1/π[1/n²-cosnx/n²-cosnx/n²+ 1/n²],
=2/πn²[1-(-1)ⁿ], because cos nx= (-1)ⁿ,
a _n =  0,if n is even,4/πn², if n is odd.
b _n =  1/π ∫_-π^π f(x)sin(nx)dx,
b _n =1/2 π[∫_-π⁰­­(π+ x) sinnx dx+1/2 π∫₀^π(π- x) sinnx  dx],
b _n =1/π[-(π+x) cos nx/n+sin nx/n²]_-π⁰+1/π[-(π-x) cos nx/n+sin nx/n²]₀^π,
b _n= 1/π[-π/n +π/n]=0.
Now substituting the values of a_0, a_n, b_n in Fourier series equation and e get,
f(x)=π/2+4/π[cosx/1²+cos3x/3²+cos5x/5²+…………..].
This is the required Fourier series of the given function.

Let f(t) be a function of t defined for 0 ≤ t≤ ∞, then the Laplace Transform of f(t) denoted by L f(t) or F(s), is defined by
L f(t) = F(s)= ^­­∫₀∞e^-stf(t)dt, provided the integral exist for all ‘s’ larger than or equal to some value ‘s₀’. The parameter s appearing in (1), is a real or Complex Number. In general ‘s’ is taken to be positive real number. The symbol L appearing in (1) is known as Laplace transform operator and the function e^-st appearing in the integral is called the kernel transform.
Now our function is f(t)= cos(at) cosh at. By the definition of Laplace transform, we have
L f(t) = F(s)= ^­­∫₀∞e^-stf(t)dt,
L f(t) = F(s)=cos(at) cosh at,
But L f(t)=L(cos(at))=s/(s²+a²)=F(s),
L f(t)=cos(at) cosh at=Lcos at(e^at+e^-at/2),
L f(t)=1/2[L(e^atcos at)+ L(e^-at cos at)],
L f(t)=1/2[F(s-a)+F(s+a)],[by first shifting theorem],
L f(t)=1/2[(s-a)s/[(s-a)²+a²]+(s+a)/ /[(s+a)²+a²],
L f(t)=1/2[(s-a)s/(s²-2as+2a²)+(s+a)/( s²+2as+2a²),
L f(t)= 2s³/(s²+2a²)²-4a²s²,
L f(t)= s³/s⁴+4a⁴.

Let f(t) be a function of ‘t’ defined for 0 ≤ t≤ ∞, then the Laplace Transform of f(t) denoted by ‘L’ f(t) or F(s), is defined by-
L f(t) = F(s)= ^­­∫₀∞e^-stf(t)dt, provided the integral exist for all ‘s’ larger than or equal to some value ‘s₀’. The parameter s appearing in (1), is a real or Complex Number. In general ‘s’ is taken to be positive real number. The symbol ‘L’ appearing in (1) is known as Laplace transform operator and the function e^-st appearing in the integral is called the kernel transform.
Now our function is f(t)= cos³2t. By the definition of Laplace transform, we have
L f(t) = F(s)= ^­­∫₀∞e^-stf(t)dt,
Since cos3t=4cos³t - 3cost,
cos³t=1/4(cos3t - 3cost),
cos³2t=1/4(cos6t - 3cos2t),
Thus, L f(t) =Lcos³2t,
Lcos³2t=L[1/4(cos6t - 3cos2t)],
Lcos³2t=1/4L(cos6t) – 3/4L(cos2t),
Lcos³2t=1/4 (s/s²+36)+3/4(s/s²+4),
Lcos³2t= s/4[1/s²+36)+3(s²+4)],
Lcos³2t= s/4[(4s²+112)/(s²+36)(s²+4)],
Lcos³2t= s(s²+28)/(s²+36)(s²+4)].

Let f(t) be a function of ‘t’ defined for 0 ≤ t≤ ∞, then the Laplace Transform of f(t) denoted by L f(t) or F(s), is defined by
L f(t) = F(s)= ^­­∫₀∞e^-stf(t)dt, provided the integral exist for all s larger than or equal to some value ‘s₀’. The parameter s appearing in (1) is a real or Complex Number. In general ‘s’ is taken to be positive real number. The symbol ‘L’ appearing in (1) is known as Laplace transform operator and the function e^-st appearing in the integral is called the kernel transform.
Let f(t)= 1√t then f(t)→∞ as t→0. Thus f(t) is not piecewise continuous in every finite interval in The Range t≥0, but still its Laplace transform exists. By the definition of Laplace transform, we have
L f(t) = F(s)= ^­­∫₀∞e^-stf(t)dt,
Therefore for f(t)= 1√t,
L1√t=∫₀∞e^-st1√tdt,
L1√t=∫₀∞e^-x x^-1/2 dx,[on putting st=x=dt=dx/s],
L1√t=1/√s∫₀∞e^-x x^(1/2)-1dx,
L1√t=1/√s_­­­­­­­­­­­­_­ gamma function (1/2), [by definition of gamma function ∫e^-uu^n-1du=gamma (n)]
L1√t=√π/s.
Thus the condition stated in Theorem 1 is not necessary for the existence of the Laplace Transform.

Let f(t) be a function of ‘t’ defined for 0 ≤ t≤ ∞, then the Laplace Transform of f(t) denoted by L f(t) or F(s), is defined by,
L f(t) = F(s)= ^­­∫₀∞e^-stf(t)dt, provided the integral exist for all ‘s’ larger than or equal to some value ‘s₀’. The parameter ‘s’ appearing in (1), is a real or Complex Number. In general ‘s’ is taken to be positive real number. The symbol ‘L’ appearing in (1) is known as Laplace transform operator and the function e^-st appearing in the integral is called the kernel transform.
Now our function is f(t)=1. By the definition of Laplace transform, we have
L f(t) = F(s)= ^­­∫₀∞e^-stf(t)dt,
Therefore for f(t)=1,
L1=∫₀∞e^-st1dt,
L1=∫₀∞e^-stdt,
L1=[e^-st/-s]₀∞, apply limit and we get,
L1=-lim_t→∞e^-st+1/s,
If s>0, then limit exists above and we obtain,
L1=1/s, s>0.