Math Examples of Integration

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 …….).

For solving differential equation first we will write the separate form.
First we write the given differential equation then we integrate equation.
 ∫2x dx = ∫(y² +1) dy,
First we integrate ‘2x’ with respect to ‘x’ and then ‘y² +1’ is integrated with respect to ‘y’
[Integration of y² is 1/3 y³ and Integration of ‘2’x is 2x²/2]
         2x²/2= 1/3 y³ + y +c,
           x² = y³/3 +y +c  [where c is integration constant ],
In this way we got the separate differential equation, which is x² = y³/3 +y +c.

For solution of separable differential equation we will use the steps shown below-
Step 1: In the first step we write the given separable differential equation
x² +5 dx + cos x cos y dy = 0,
Step 2: After first step we are separated the given differential equation
x² +5 dx + sec x cos y dy = 0,
 cos y dy = -(x²+5)cos x dx,
We assume: LHS = cos y dy,   RHS = -(x²+5)cos x dx.
Step 3: In this step we have integrated both sides.
Now first we will integrate left hand side part
In this part we integrate the equation with respect to ‘y’,
∫cos y dy = sin y,
Now integrate right-hand side, with respect to ‘x’.
∫ -(x²+5)cos x dx = -∫(x+5)cos x dx,
                   = -[(x+5)sin x - ∫ sin x dx],
                  = -[(x+5) sin x + cos x + c],  
Here ‘c’ is Integration constant.
  Step 4: In this step we are putting the values which are solved above.
 Plugging the value both sides,
sin y = -[(x+5) sin x + cos x + c],
In this way we get the separable differential equation which is
sin y = -[(x+5) sin x + cos x + c].

In the first step we write the given separable differential equation,
y dy + sec x  tan y dx = 0,
Step 2: After first step we have separated the given differential equation
sec x tan y dx = -y dy,
 sec x dx = -y cot y dy,
We assume:   LHS = sec x dx  ,   RHS = -y cot y dy,
Step 3: In this step we have integrated both sides.
Now first we integrate left hand side part,
In this part we integrate the equation with respect to ‘x’,
∫sec x dx = ln|tan x + sec x| + c,
Now we will integrate the right-hand side part with respect to ‘y’.
∫ -y cot y dy = - ∫ y cot y dy,
                   = - [y ln|sin y| - ∫ ln |sin y| dy],
                   = -(y ln|sin y| + sin y ln |sin y| - sin y),
[by ln cx= x ln cx – x].
Step 4: In this step we are putting the value which are solved above.
 Plugging the value in both sides,
ln|tan x + sec x| + c  =  -(y ln|sin y| + sin y ln |sin y| - sin y)  
where ‘c’ is Integration constant of this equation.
In this way we get the separable differential equation which is
ln|tan x + sec x| + c  =  -(y ln|sin y| + sin y ln |sin y| - sin y).  

For solution of separable differential equation we will use steps shown below-
 Step 1: In first step we write the given separable differential equation
  x dx + sin x cos y dy = 0,
Step 2: After first step we will separate the given differential equation,
sec x cos y dy = -x dx,
 cos y dy = -x cos x dx,
We assume:   LHS = cos y dy,   RHS = -x cos x dx,
Step 3: In this step we are integrated the both side.
Now first we will integrate left hand side part
In this part we integrate the equation with respect to ‘y’,
∫cos y dy = sin y,
No  we will integrate the right-hand side part with respect to ‘x’.
∫ -x cos x dx = - ∫ x cos x dx,
                     = - [x sin x - ∫ sin x dx],
                    = -(x sin x + cos x + c).
Step 4: In this step we are putting the value which are solved above.
 Plugging the value in both sides-
sin y  =  -(x sin x + cos x + c),
sin y = x sin x + cos x + c, 
where ‘c’ is Integration constant of this equation.
In this way we get the separable differential equation which is
sin y = x sin x + cos x + c.

For solving Antiderivative we need to follow the steps shown below:
Step 1: In the first step we write the given function.
f(x) = x⁴ +cot x,
Step 2: Now we integrate the both side of the function,
∫f(x) dx = ∫ x⁴ +cot x dx,
Step 3: In this step we will separate the integral function.
∫(x⁴ +cot x) dx = ∫x⁴ dx + ∫cot x dx, 
Step 4: After above step we will integrate each function with respect to ‘x’.
∫(x⁴ +cot x) dx = x⁵/5 + ln|sin x| +c [Here x⁴ integration is x⁵/5 and Integration of cot x is ln|sin x|]
(Where ‘c’ is integration constant),
At last we get the antiderivative of given function
x⁵/5 + ln|sin x| +c.

For solving Antiderivative we need to follow the steps shown below:
Step 1: In the first step we write the given function.
f(x)= 3x⁴ +6x² +7x +1,
Step 2: Now we integrate the both side of the function,
∫f(x) dx = ∫3x⁴ + 6x² +7x +1  dx,
Step 3: In this step we separate the integral function.
∫(3x⁴ + 6x² +7x+ 1) dx = ∫3x⁴ dx + ∫6x² dx + ∫7x dx + ∫1 dx,
Step 4: After above step we will integrate each function with respect to ‘x’.
∫(3x⁴ + 6x² +7x +1) dx = 3x⁵/5 + 6x³/3 + 7x²/2+ x +c [Here x⁴ integration is x⁵/5 ,integration of x² is x³/3 , integration of x is x²/2 and Integration 1 is  ‘x’].
(Where ‘c’ is integration constant).
Now we got the antiderivative of given function
3x⁵/5 + 6x³/3 + 7x²/2+ x +c.

For solving Antiderivative we need to follow the steps shown below:
Step 1: In the first step we write the given function.
f(x)= tan x +6x + tan²x,
Step 2: Now we will integrate both sides of the function
∫f(x) dx = ∫tan x + 6x +tan²x dx.
Step 3: In this step we will separate the integral function.
∫( tan x +6x + tan²x) dx = ∫tan x dx + ∫6x dx + ∫tan²x dx,
Step 4: After above step we will integrate each function with respect to ‘x’.
∫( tan x +6x + tan²x)dx = -ln|cos x| + 6x²/2 + tan x - x +c [Here tan x Integration is –ln|cos x| and integration of x is x²/2 and integration of tan²x is tan x - x].
(Where ‘c’ is integration constant).
Now we got the antiderivative of given function,
-ln|cos x| + 6x²/2 + tan x - x +c. 

For solving Antiderivative we need to follow the steps shown below:
Step 1: In the first step we write the given function.
f(x)= sin x +cos x + x,
Step 2: Now we integrate the both side of the function
∫f(x) dx = ∫(sin x + cos x +x)dx,
Step 3: In this step we will separate the integral function.
∫(sin x + cos x +x) dx = ∫sin x dx + ∫cos x dx + ∫x dx
Step 4: After above step we will integrate each function with respect to ‘x’.
∫(sin x+ cos x +x) dx = -cos x + sin x + x²/2 +c [Here sin x Integration is –cos x and integration of cos x is sin x and integration of x is x²/2].
(Where ‘c’ is integration constant).
At last we got the antiderivative of given function
-cos x + sin x + x²/2 +c.

For solving Antiderivative we need to follow the steps shown below:
Step: In the first step we write the given function.
f(x)= x⁴ + sin x +1,
Step 2: Now we integrate the both side of the function
∫f(x) dx = ∫x⁴ + sin x +1 dx,
Step 3: In this step we separate the integral function.
∫(x⁴ + sin x +1) dx = ∫x⁴ dx + ∫sin x dx + ∫1 dx,
Step 4: After above steps we will integrate each function with respect to ‘x’.
∫(x⁴ + sin x +1) dx = x⁵/5 – cos x + x +c [Here x⁴ integration is x⁵/5, and Integration of sin x is –cos x and integration of 1 is ‘x’].
(Where ‘c’ is integration constant)
At last we get the antiderivative of given function,
x⁵/5 –cos x + x +c.