∫∫e^x^y dydx

∫∫e^x^y dydx
= ∫∫Σ(n = 0 to ∞)[(x^y)ⁿ/n!] dydx
= ∫∫[1/n! + Σ(n = 1 to ∞)[(x^(ny))/n!]] dydx
= ∫[y/n! + Σ(n = 1 to ∞)[(x^(ny))/(n(n!)ln x)] + C_1] dx

Let u = ln x
x = e^u
dx = e^u du

∴∫[y/n! + Σ(n = 1 to ∞)[(x^(ny))/(n(n!)ln x)] + C_1] dx
= ∫[y/n! + Σ(n = 1 to ∞)[((e^u)^(ny))/(n(n!)u)] + C_1]e^u du
= ∫[(ye^u)/n! + (e^u)Σ(n = 1 to ∞)[(e^(nyu))/(n(n!)u)] + (C_1)e^u] du
= ∫[(ye^u)/n! + Σ(n = 1 to ∞)[(e^((ny + 1)u))/(n(n!)u)] + (C_1)e^u] du
= ∫[(ye^u)/n! + Σ(n = 1 to ∞)[Σ(k = 0 to ∞)[(((ny + 1)u)^k)/(n(n!)(k!)u)]] + (C_1)e^u] du
= ∫[(ye^u)/n! +Σ(n = 1 to ∞)[1/(n(n!)u)] + Σ(n = 1 to ∞)[Σ(k = 1 to ∞)[((ny + 1)^k u^(k – 1))/(n(n!)(k!))]] + (C_1)e^u] du
= (ye^u)/n! +Σ(n = 1 to ∞)[(ln│u│)/(n(n!))] + Σ(n = 1 to ∞)[Σ(k = 1 to ∞)[((ny + 1)^k u^k)/(nk(n!)(k!))]] + (C_1)e^u + C_2
= (xy)/n! +Σ(n = 1 to ∞)[(ln│ln x│)/(n(n!))] + Σ(n = 1 to ∞)[Σ(k = 1 to ∞)[((ny + 1)^k (ln x)^k)/(nk(n!)(k!))]] + (C_1)x + C_2

2007-07-21 15:17:23 補充：
唔好意思，打錯少少嘢！請把每行的首項中分母的n!除去。

2007-07-22 01:40:06 補充：
唔好意思，計錯數！重新再計過！

2007-07-22 01:44:09 補充：
Evaluate ∫∫e^(x^y) dydx.∫∫e^(x^y) dydx= ∫∫Σ(n = 0 to ∞)[(x^y)ⁿ/n!] dydx（use e^a = Σ(n = 0 to ∞)[aⁿ/n!] and put a = x^y）= ∫∫[1 十 Σ(n = 1 to ∞)[(x^(ny))/n!]] dydx= ∫∫[1 十 Σ(n = 1 to ∞)[(e^(ln (x^(ny))))/n!]] dydx（x^c = e^(ln (x^c))）= ∫∫[1 十 Σ(n = 1 to ∞)[(e^((n ln x)y))/n!]] dydx

2007-07-22 01:46:34 補充：
= ∫[y 十 Σ(n = 1 to ∞)[(e^((n ln x)y))/(n(n!)ln x)] 十 C_1] dx= ∫[y 十 Σ(n = 1 to ∞)[(e^(ln (x^(ny))))/(n(n!)ln x)] 十 C_1] dx= ∫[y 十 Σ(n = 1 to ∞)[(x^(ny))/(n(n!)ln x)] 十 C_1] dxLet u = ln xx = e^udx = e^u du

2007-07-22 01:48:32 補充：
∴∫[y 十 Σ(n = 1 to ∞)[(x^(ny))/(n(n!)ln x)] 十 C_1] dx= ∫[y 十 Σ(n = 1 to ∞)[((e^u)^(ny))/(n(n!)u)] 十 C_1]e^u du= ∫[ye^u 十 (e^u)Σ(n = 1 to ∞)[(e^(nyu))/(n(n!)u)] 十 (C_1)e^u] du= ∫[ye^u 十 Σ(n = 1 to ∞)[(e^((ny 十 1)u))/(n(n!)u)] 十 (C_1)e^u] du

2007-07-22 01:50:26 補充：
= ∫[ye^u 十 Σ(n = 1 to ∞)[Σ(k = 0 to ∞)[(((ny 十 1)u)^k)/(n(n!)(k!)u)]] 十 (C_1)e^u] du（use e^b = Σ(k = 0 to ∞)[(b^k)/k!] and put b = (ny 十 1)u）= ∫[ye^u 十 Σ(n = 1 to ∞)[1/(n(n!)u)] 十 Σ(n = 1 to ∞)[Σ(k = 1 to ∞)[((ny 十 1)^k u^(k – 1))/(n(n!)(k!))]] 十 (C_1)e^u] du

2007-07-22 01:51:19 補充：
= ye^u 十 Σ(n = 1 to ∞)[(ln│u│)/(n(n!))] 十 Σ(n = 1 to ∞)[Σ(k = 1 to ∞)[((ny 十 1)^k u^k)/(nk(n!)(k!))]] 十 (C_1)e^u 十 C_2= xy 十 Σ(n = 1 to ∞)[(ln│ln x│)/(n(n!))] 十 Σ(n = 1 to ∞)[Σ(k = 1 to ∞)[((ny 十 1)^k (ln x)^k)/(nk(n!)(k!))]] 十 (C_1)x 十 C_2

2007-07-22 01:54:25 補充：
最上面果大篇大字你當佢垃圾就算啦！

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