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Question Number 169259 by mathlove last updated on 27/Apr/22

Commented by infinityaction last updated on 27/Apr/22

I(a) = ∫_0 ^∞ (e^(−ax) /x) dx and I(b) = ∫_0 ^∞ (e^(−bx) /x)dx I′(a) = −∫_0 ^∞ e^(−ax) dx , I′(b) = −∫_0 ^∞ e^(−bx) dx I′(a) = [(e^(−ax) /a)]_0 ^∞ , I′(b) = [(e^(−bx) /b)]_0 ^∞ I′(a)= −(1/a) , I′(b) =− (1/b) I(a) = −log ∣a∣ , I(b) = −log ∣b∣ I(a)−I(b) = log ∣b∣−log ∣a∣ I = log ∣(b/a)∣

I(a)=∫0∞e−axxdxandI(b)=∫0∞e−bxxdxI′(a)=−∫0∞e−axdx,I′(b)=−∫0∞e−bxdxI′(a)=[e−axa]0∞,I′(b)=[e−bxb]0∞I′(a)=−1a,I′(b)=−1bI(a)=−log∣a∣,I(b)=−log∣b∣I(a)−I(b)=log∣b∣−log∣a∣I=log∣ba∣

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