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)∣](https://www.tinkutara.com/question/Q169267.png)
$$\:\:\:\:\:\:\:\:{I}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{ax}} }{{x}}\:{dx}\:\:\:\:\:\:{and}\:\:\:\:{I}\left({b}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{bx}} }{{x}}{dx} \\ $$$$\:\:\:\:\:\:{I}'\left({a}\right)\:\:=\:\:−\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} {dx}\:\:\:\:,\:\:\:{I}'\left({b}\right)\:=\:−\int_{\mathrm{0}} ^{\infty} {e}^{−{bx}} {dx} \\ $$$$\:\:\:\:\:\:{I}'\left({a}\right)\:=\:\left[\frac{{e}^{−{ax}} }{{a}}\right]_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\:\:\:\:,\:\:\:{I}'\left({b}\right)\:=\:\left[\frac{{e}^{−{bx}} }{{b}}\right]_{\mathrm{0}} ^{\infty} \\ $$$$\:\:\:\:\:\:{I}'\left({a}\right)=\:−\frac{\mathrm{1}}{{a}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:,\:\:\:{I}'\left({b}\right)\:=−\:\frac{\mathrm{1}}{{b}} \\ $$$$\:\:\:\:\:{I}\left({a}\right)\:=\:−\mathrm{log}\:\mid{a}\mid\:\:\:\:\:,\:\:\:{I}\left({b}\right)\:=\:−\mathrm{log}\:\mid{b}\mid \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:{I}\left({a}\right)−{I}\left({b}\right)\:=\:\mathrm{log}\:\mid{b}\mid−\mathrm{log}\:\mid{a}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{I}\:\:=\:\:\mathrm{log}\:\mid\frac{{b}}{{a}}\mid \\ $$