How to express F(x) in terms of G(x)?

Question

How to express $$F(x)=\int _{ 1 }^{ x }{ \frac { e^{ t } }{ t^2 } dt} $$ in terms of $$ G(x)=\int _{ 1 }^{ x }{ \frac { e^{ t } }{ t } dt} $$ ?

Have you tried integration by parts? – Alex R. Nov 23 '15 at 21:13 — Alex R., Nov 23 '15 at 21:13

Balloon · Accepted Answer · 2015-11-23T21:30:31.680

1

You have, taking $u=-\frac{1}{t}$ (and so $u'=\frac{1}{t^2}$), $v=e^t$ and using the integration by parts formula $(\int u'v=[uv]-\int uv')$ that : $$F(x)=\int_1^x\frac{e^t}{t^2}dt=\Big[-\frac{e^t}{t}\Big]_1^x+\int_1^x\frac{e^t}{t}dt,$$ and so $$F(x)=e-\frac{e^x}{x}+G(x).$$

edited Nov 23 '15 at 21:30

answered Nov 23 '15 at 21:16

Balloon

8,374

I didnt get the fist step.What is your first function and what is your second function during by parts? – Nov 23 '15 at 21:19
@SanchayanDutta : I edited that ! – Balloon Nov 23 '15 at 21:30
@Baloown This is wrong. – Prabath Hewasundara Nov 23 '15 at 21:35
@RPH : Yes I edited that by the way too, thank you ! – Balloon Nov 23 '15 at 21:35
@Baloown ok..(y) – Prabath Hewasundara Nov 23 '15 at 21:37

Prabath Hewasundara · Answer 2 · 2015-11-23T21:33:31.193

0

Use integration by parts

$$F(x)=\int_1^x\frac{e^t}{t^2}dt=\int_1^xe^t\cdot(-1)(\frac{d}{dt}(\frac{1}{t}))dt=(\frac{-e^t}{t})_1^x+\int_1^x\frac{e^t}{t}dt=\frac{-e^x}{x}+e+G(x)$$

$$F(x)=G(x)+e-\frac{e^x}{x}$$

edited Nov 23 '15 at 21:33

answered Nov 23 '15 at 21:27

Prabath Hewasundara

1,644

How to express F(x) in terms of G(x)?

2 Answers2