# Integral with exponential function

Integrate[((Sqrt[g*(x + h)/y])^V) * Exp[-u * Sqrt[g*(x + h)/y]/n] * Exp[-L*x], {x, 0, Infinity}]

• if you can integrate $\int x^n e^{-x^2-x} \, dx$ then you might have a chance of integrating your integral. Not all integrals can be integrated. Jun 16, 2021 at 13:48

I assumed the parameters are positive real numbers. One could simplify things a little by replacing g/y by a and u/n by b — the more parameters, the harder you make it for Mathematica.

Integrate evaluates in the special case h = 0, but not for other values of h tested.

Clear[g, L, x, h, y, V, n, u];
Block[{h = 0},
Integrate[((Sqrt[g*(x + h)/y])^V)*Exp[-u*Sqrt[g*(x + h)/y]/n]*
Exp[-L*x], {x, 0, Infinity},
Assumptions -> V > 0 && g > 0 && u > 0 && n > 0 && L > 0 && y > 0]
]
(*
(2^(-1 - V) ((L y)/g)^(-V/2)
Gamma[2 + V] HypergeometricU[1 + V/2, 1/2, (g u^2)/(4 L n^2 y)])/L
*)


It evaluates if only V is given numeric values (in all tests):

Block[{g, h, V = 3, L, n, u, y},
Integrate[((Sqrt[g*(x + h)/y])^V)*Exp[-u*Sqrt[g*(x + h)/y]/n]*
Exp[-L*x], {x, 0, Infinity},
Assumptions ->
V > 0 && g > 0 && u > 0 && n > 0 && L > 0 && y > 0 && h > 0]
]


The remaining cases probably cannot be done.

• This is strange. Why does Mathematica integrate the definite integral but not the indefinite one? !Mathematica graphics I am using V 12.3. I always start with the indefinite case, and if that works, only then I try the definite integral. Jun 16, 2021 at 14:56
• @Nasser There are methods available for definite integrals, for example, contour integration, that are not available for indefinite ones. Jun 16, 2021 at 15:02