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

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

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

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

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

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

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

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

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

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


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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What makes you think any expression simpler than $h(x)=f(x,1)$ exists? – Did Nov 1 '12 at 9:16
@did $\int_{0}^{\infty} \frac{1}{x+e^{y t}} dt=\frac{\ln(1+x)}{xy}$. I just wanted to go more step to add $t$ next to $x$ and then I noticed that $h(x)$ is very beautiful series in my opinion. I just wanted to search about it. I thought maybe someone knows about it. Now I have been tring to find a differential equation that satisfies $h(x)$ and then to express it as known functions. Thanks – Mathlover Nov 1 '12 at 9:29
Why is it one more step? One more step on what path, leading to what goal? – Did Nov 1 '12 at 9:32
@did I knew the result $\int_{0}^{\infty} \frac{1}{x+e^{y t}} dt=\frac{\ln(1+x)}{xy}$. I just wanted to add $t$ next to $x$ and then to see the result. It is just my personal curiosity. – Mathlover Nov 1 '12 at 9:36
@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. – doraemonpaul Mar 26 '14 at 5:04

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