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I have the following question:

if $1=\int _{ 1 }^{ \infty }{ \frac { ax+b }{ x(2x+b) } dx } $ then $a+b=?$

a) $0$

b) $e$

c) $2e-2$

d) $1$

I tried finding it's anti-derivative but that doesn't seem to help.

I also tried to get it to $\int _{ 1 }^{ \infty }{(\alpha-1) \frac { 1 }{ x^\alpha } dx } $ with $\alpha>1$ since that equals $1$ but without finding an answer.

What am I missing here?

Thanks.

Community
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Paz
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  • You tried finding the antiderivative. Did you succeed in finding the antiderivative? – Gerry Myerson Jun 25 '13 at 12:10
  • you can try integration by parts. – dajoker Jun 25 '13 at 12:11
  • yes: $1=\int { 1 }^{ \infty }{ \frac { 1 }{ x } dx }+ \int _{ 1 }^{ \infty }{ \frac { a-2 }{ 2x+b } dx }={ \left[ \ln { (x) } \right] }{ 1 }^{ \infty }+\frac { a }{ 2 } { \left[ \ln { (2x+b) } \right] }{ 1 }^{ \infty }-{ \left[ \ln { (2x+b) } \right] }{ 1 }^{ \infty }$ – Paz Jun 25 '13 at 12:15
  • @anorton if you add the two you get the original integral. Is there anything wrong there? – Paz Jun 25 '13 at 12:45
  • @Paz Oops... never mind. (didn't add correctly) – apnorton Jun 25 '13 at 12:54

2 Answers2

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Hints: (1) If $a\ne 0$ what kind of trouble do we get into? (2) Partial fractions.

André Nicolas
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Note that, for the integral to converge, $a$ has to equal $0$ which implies

$$ 1=\int _{ 1 }^{ \infty }{ \frac { b }{ x(2x+b) } dx }=\ln(b+2)-\ln(2)=\ln(\frac{b+2}{2}) $$

$$ \implies \frac{b+2}{2}=e \implies b=2 e -2 \implies a+b= 2 e-2 .$$

Note: Observe that, the integrand

$$ \frac { ax+b }{ x(2x+b) }\sim \frac{ax}{2x^2}=\frac{a}{2x}, $$

which is not integrable on $[1,\infty)$.

Mhenni Benghorbal
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