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enter image description here I was trying to solve this question but got stuck. If we further solve it we get $I'(e)=0 $ which does no help to find the value of the integral. I know an alternate way to take $7$ as variable but I want to know if the process I used can help in anyway to evaluate this integral. How will we find $I(e)$ if $I'(e)$ comes 0 here? To be more clear I want to know how to evaluate this question using Leibnitz rule and differentiating with respect to $e$. I finally got $ I(e)=0+c$ Further I tried to find $ I(1) $ and I found it's giving 0 that would imply my $ c=0 $ that means answer should be $0$ but it's $ ln7 $. Please correct me where am I wrong.

Jasmine
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1 Answers1

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Using Double Integration

Let $\displaystyle I = \int^{\infty}_{0}\frac{e^{-x}-e^{-7x}}{x}dx = \int^{\infty}_{0}\bigg(\int^{7}_{1}e^{-tx}dt\bigg)dx$

So $\displaystyle I = \int^{7}_{1}\bigg(\int^{\infty}_{0}e^{-xt}dx\bigg)dt = \int^{7}_{1}\frac{1}{t}dt=\ln(t)\bigg|^{7}_{1} = \ln(7).$

Using Differentiation under integral sign

Let $\displaystyle I(\alpha) = \int^{\infty}_{0}\frac{e^{-\alpha x}-e^{-x}}{x}dx,$ Then $\displaystyle I'(\alpha) = -\int^{\infty}_{0}\frac{x\cdot e^{-\alpha x}}{x}dx = -\frac{e^{-\alpha x}}{\alpha}\bigg|^{\infty}_{0}=\frac{1}{\alpha}$

So $\displaystyle I (\alpha) = \ln(\alpha)+\mathcal{C}.$ Put $\alpha =1,$ We get $I(1)=0$

So we gave $\displaystyle I (\alpha) = \int^{\infty}_{0}\frac{e^{-\alpha x}-e^{-x}}{x}dx= \ln(\alpha).$

So $\displaystyle I(7)=\int^{\infty}_{0}\frac{e^{-7x}-e^{-x}}{x}dx = \ln(7).$

DXT
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