Is it possible to solve this equation using lambert function

Question

Can the following equation be solved using Lambert function?

$$x(1+e^x)=a$$

Answer 1

We know that mixed exponential/bilinear equations can be solved by the extended Lambert function $W_r(x)$, which can be represented by the following Lagrange inverting series:

$$z(A,t,s)=t- (t-s) \sum_{n=1} \frac{L_n' (n(t-s))}{n} e^{-nt} A^n$$

which is the solution of:

$$e^z=A\frac{z-t}{z-s}$$

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References

On the generalization of the Lambert $W$ function with applications in theoretical physics, http://arxiv.org/abs/1408.3999

[68] C. E. Siewert and E. E. Burniston, "Solutions of the Equation $ze^z=a(z+b)$," Journal of Mathematical Analysis and Applications, 46 (1974) 329-337. http://www4.ncsu.edu/~ces/pdfversions/68.pdf

Answer 2

$\def\M{\operatorname M}$ Instead of Lagrange inversion, let’s apply Mellin inversion:

$$f(x)=x(e^x+1)\implies f^{-1}(x)=h(x)=\M^{-1}_x(\M_s(h(t))\\\M_s(h(t))=\int_0^\infty t^{s-1}h(t)dt=\int_0^\infty t f(t)^{s-1}df(t)$$

Next, use integration by parts and the $(z+1)^a$ Mellin Barnes integral to get

$$\M_s(h(t))=\frac1s\int_0^\infty (t(e^t+1))^sdt=-\frac i{2\pi s \Gamma(-s)}\int_{c-i\infty}^{c+i\infty} \Gamma(-s-w)\Gamma(w)\int_0^\infty t^se^{-wx}dtdw$$

The $t$ integral is a gamma function and the resulting $s$ inverse Mellin transform integral gives confluent hypergeometric function. Therefore:

$$\bbox[4px,border:3px solid #1ae]{f(x)=x(e^x+1)\implies f^{-1}(x)=-\frac 1{4\pi^2}\int_{c_1-i\infty}^{c_1+i\infty}\int_{c_2-i\infty}^{c_2+i\infty}\frac{\Gamma(-s-w)\Gamma(w)s!}{x^sw^{s+1}(-s)!} dsdw=-\frac{ix}2\int_{c-i\infty}^{c+i\infty}\csc(\pi t)\,_1\text F_1(1-t;2;-tx)dt=\frac x{2\pi}\int_0^{2\pi}e^{it}\ln\left(e^{xe^{it}}(e^{-i t}+1)+1\right)dt}$$

where approximately $0<c_1,-c_2<1$. Also, converting the $_1\text F_1$ function into a Laguerre $\operatorname L_{t-1}^1(w)$ function and using its integral representation gives the final result.

All verified here:

Answer 3

No. Lambert's W function solves only cases of the form $~xe^x=$ constant, whereas here we have $xe^x=$ variable.