Help needed in proving a integral equation

[vineel49]

Homework Statement

Help needed in simplifying this one
$$\left[\int\limits_0^{Inf} {\int\limits_0^{Inf} {{e^{ - \alpha X - \beta Y}} \cdot F(X + Y + c)} } \cdot {X^d} \cdot {Y^e} \cdot dX \cdot dY\right]$$ is equal to

$$\left[\sum\limits_{i = 0}^d {d{C_k} \cdot } {( - 1)^{d - i}} \cdot \left[ {\left( {\frac{{factorial(d + e - i)}}{{{{(\beta - \alpha )}^{d + e - i + 1}}}} \cdot \{ \int\limits_0^{Inf} {{\operatorname{e} ^{\alpha t}} \cdot F(t + c) \cdot {t^i} \cdot dt)} \} } \right) - \left( {\sum\limits_{j = 0}^{d + e - i} {\frac{{S(d + e - i,j - 1)}}{{{{(\beta - \alpha )}^{j + 1}}}}} \cdot \{ \int\limits_0^{Inf} {{\operatorname{e} ^{\beta t}} \cdot F(t + c) \cdot {t^{d + e - j}} \cdot dt)} \} } \right)} \right]\right]$$

Homework Equations

alpha,beta are constants; d & e are non negative integers, X & Y are variables
dCk is binomial coefficient. S(x1,x2)=x1*(x1-1)*...(x1-x2) and S(x1,x2)=1 when x2<0. F is a even function

The Attempt at a Solution

put X+Y=V and Y=U. All I need is help in start up -JUST FIRST 2-3 steps, from there I can solve on my own. After taking X+Y=V and Y=U , how many integrals will come?
Is this the right way to start
$$\left[\int\limits_0^{Inf} {\int\limits_U^{Inf} {{e^{ - \alpha (V - U) - \beta (U)}} \cdot F(V + c) \cdot } } {(V - U)^d} \cdot {U^e} \cdot dV \cdot dU\right]$$ ?
There is something wrong in
$$\left[\int\limits_0^{Inf} {\int\limits_U^{Inf} {( - )dV \cdot dU} } \right]$$

 

[vineel49]

I started simplifying from
$$\left[\int\limits_0^{Inf} {\int\limits_U^{Inf} {{e^{ - \alpha (V - U) - \beta (U)}} \cdot F(V + c) \cdot } } {(V - U)^d} \cdot {U^e} \cdot dV \cdot dU\right]$$ . Finally I coudn't reach right hand side of the equation. There is something wrong in
$$\left[\int\limits_0^{Inf} {\int\limits_U^{Inf} {( - )dV \cdot dU} } \right]$$

 

[tiny-tim]

hi vineel49!

Is this the right way to start
$$\left[\int\limits_0^{Inf} {\int\limits_U^{Inf} {{e^{ - \alpha (V - U) - \beta (U)}} \cdot F(V + c) \cdot } } {(V - U)^d} \cdot {U^e} \cdot dV \cdot dU\right]$$

yes

now use the binomial theorem on the (V - U)^d,

then your ∫∫ is the sum of lots of separable ∫∫, of the form

∫∫ f(U)g(V) dUdV = {∫ f(U) dU} {∫ g(V) dV}

 

[vineel49]

hi vineel49!


yes

now use the binomial theorem on the (V - U)^d,

then your ∫∫ is the sum of lots of separable ∫∫, of the form

∫∫ f(U)g(V) dUdV = {∫ f(U) dU} {∫ g(V) dV}

I did in the way u suggested, but I was left out with a 'U' variable. but the answer doesnot contain 'U' .

 

[tiny-tim]

I did in the way u suggested, but I was left out with a 'U' variable. but the answer doesnot contain 'U' .

yes, but you can integrate the {∫ f(U) dU} (which of course eliminates U), since that function f is known