※ 引述《ElvinN (Elvin)》之銘言：
: ※ 引述《k6416337 (第一次獻給了涼宮版-光希)》之銘言：
: : 1.
: : ∞
: : ∫ (sin8x)^2-(sin2x)^2 dx
: : 0 ￣￣￣￣￣x￣￣￣￣￣
: : 它有提示說要用FUBINI定理，可是我不知道要怎麼用= =
: 1.應該不是很難的題目，先用(sinx)^2的公式轉換後，很快答案就出來了
Well, at least for me, it's far from trivial....
If you don't know the Laplace transform, and the relation of
1/t * exp(-at), it could be very hard to do it.
Here is another way of dealing this. Use Fubini theorem.
In the end, you can simpify the sin(ax)^2 into exp() form,
So we only consider the integration
1/x * exp(-ax), x from 0 to inf.
The trick part is, if we write exp(-ax) as the integration
exp(-ax) = Integral of u [a, inf] , -x*exp(-ux)
Then you will have a double integral, but x can cancel out the 1/x
Change the order of integration, done.
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