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$x,y>0$ $f(x,y)=\int_{0}^{\infty} \frac{1}{xt+e^{y t}} dt$

if $x=0$ then $f(0,y)=1/y$

$f(x,y)=\int_{0}^{\infty} \frac{1}{e^{y t}(1+xte^{-y t})} dt=\int_{0}^{\infty} \frac{e^{-y t}}{(1+xte^{-y t})} dt$

$f(x,y)=\int_{0}^{\infty} e^{-y t}(1-xte^{-y t}+x^2t^2e^{-2y t}-x^3t^3e^{-3y t}+....) dt$

$f(x,y)=\int_{0}^{\infty} e^{-y t}\sum_{k=0}^{\infty} (-1)^kx^kt^ke^{-ky t} dt$

$f(x,y)=\int_{0}^{\infty} \sum_{k=0}^{\infty} (-1)^kx^kt^ke^{-(k+1)y t} dt$

$f(x,y)=\sum_{k=0}^{\infty}\bigg((-1)^kx^k\int_{0}^{\infty} t^ke^{-(k+1)y t} dt\bigg)$

$g(y)=\int_{0}^{\infty} t^ke^{-(k+1)y t} dt=\frac{k!}{(k+1)^{(k+1)}y^{(k+1)}}$

$f(x,y)=\sum_{k=0}^{\infty}\frac{(-1)^kx^k k!}{(k+1)^{(k+1)}y^{(k+1)}}=\frac{1}{y}\sum_{k=0}^{\infty}\frac{(-1)^k k!}{(k+1)^{(k+1)}}(\frac{x}{y})^k$

$h(x)=\sum_{k=0}^{\infty}\frac{(-1)^k k!}{(k+1)^{(k+1)}}x^k$

$f(x,y)=\frac{1}{y}h(\frac{x}{y})$

What is the closed form of $h(x)$ as known functions?

Is there a method to evaluate $\int_{0}^{\infty} \frac{1}{xt+e^{y t}} dt$? I tried different kind of variable change strategy but it did not work.

Thanks a lot for answers

EDIT: Added as additional information:

$\int_{0}^{\infty} \frac{1}{xt^n+e^{y t}} dt=\sum_{k=0}^{\infty}\bigg((-1)^kx^k\int_{0}^{\infty} t^{kn}e^{-(k+1)y t} dt\bigg)=$

$=\sum_{k=0}^{\infty}\frac{(-1)^kx^k (nk)!}{(k+1)^{(nk+1)}y^{(nk+1)}}$

if $n=0$ then

$\int_{0}^{\infty} \frac{1}{x+e^{y t}} dt=\sum_{k=0}^{\infty}\frac{(-1)^kx^k }{(k+1)y}=\frac{1}{xy}\sum_{k=0}^{\infty}\frac{(-1)^kx^{k+1} }{(k+1)}=\frac{\ln(x+1)}{xy}$

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    @Mathlover: You basically did this correctly, but you should beware that for some ratios between $x$ and $y$ , the maximum value of $xte^{-yt}$ will exceed $1$ , at this time you should need another method.2014-03-26

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