1.
f(x)=−∞∞∫e−2x(1−e−x)2dx
Changing x with -x we get;
f(−x)=−∞∞∫e2x(1−ex)2dx
Now f(x)+f(−x)=0
And also f(x)−f(−x)=0
Thus, f(x) is neither even nor odd.
2. Let g(x)=−∞∞∫xe−2x(1−e−x)2dx
Changing x with -x, we get;
g(−x)=−∞∞∫−xe2x(1−ex)2dx
Now, g(x)+g(−x)=0
And also g(x)−g(−x)=0
Thus, g(x) is also neither even nor odd.
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