Evaluating an improper integral using double integrals [duplicate]

1

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This question already has an answer here:

• 
  Best way to integrate $ int_0^infty frac{e^{-at} - e^{-bt}}{t} text{d}t $
  
  2 answers
  
  

working on a problem to evaluate

$int_0^infty frac {e^{-x} - e^{-ax}} {x} dx$

the instructions say to first evaluate
$int_1^a e^{-xy} dy$

which comes out to the integrand of the original improper integral.

using this it seems we can rewrite the improper integral as the double integral as follows

$int{_0^infty}int{_1^a} e^{-xy}dydx$

However where I am stuck is changing the order of integration on this double integral in order to evaluate it.

integration improper-integrals order-of-integration

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asked Jan 2 at 23:59

heironymousheironymous

41

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marked as duplicate by Zacky, RRL, Lord Shark the Unknown, KReiser, Community♦ Jan 3 at 4:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See: en.wikipedia.org/wiki/Frullani_integral or set $a=1$ here:math.stackexchange.com/q/62254/515527
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  – Zacky
  Jan 3 at 0:01
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

This question already has an answer here:

• 
  Best way to integrate $ int_0^infty frac{e^{-at} - e^{-bt}}{t} text{d}t $
  
  2 answers
  
  

working on a problem to evaluate

$int_0^infty frac {e^{-x} - e^{-ax}} {x} dx$

the instructions say to first evaluate
$int_1^a e^{-xy} dy$

which comes out to the integrand of the original improper integral.

using this it seems we can rewrite the improper integral as the double integral as follows

$int{_0^infty}int{_1^a} e^{-xy}dydx$

However where I am stuck is changing the order of integration on this double integral in order to evaluate it.

integration improper-integrals order-of-integration

share|cite|improve this question

asked Jan 2 at 23:59

heironymousheironymous

41

$endgroup$

marked as duplicate by Zacky, RRL, Lord Shark the Unknown, KReiser, Community♦ Jan 3 at 4:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See: en.wikipedia.org/wiki/Frullani_integral or set $a=1$ here:math.stackexchange.com/q/62254/515527
  $endgroup$
  – Zacky
  Jan 3 at 0:01
  
  
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

This question already has an answer here:

• 
  Best way to integrate $ int_0^infty frac{e^{-at} - e^{-bt}}{t} text{d}t $
  
  2 answers
  
  

working on a problem to evaluate

$int_0^infty frac {e^{-x} - e^{-ax}} {x} dx$

the instructions say to first evaluate
$int_1^a e^{-xy} dy$

which comes out to the integrand of the original improper integral.

using this it seems we can rewrite the improper integral as the double integral as follows

$int{_0^infty}int{_1^a} e^{-xy}dydx$

However where I am stuck is changing the order of integration on this double integral in order to evaluate it.

integration improper-integrals order-of-integration

share|cite|improve this question

asked Jan 2 at 23:59

heironymousheironymous

41

$endgroup$

This question already has an answer here:

• 
  Best way to integrate $ int_0^infty frac{e^{-at} - e^{-bt}}{t} text{d}t $
  
  2 answers
  
  

working on a problem to evaluate

$int_0^infty frac {e^{-x} - e^{-ax}} {x} dx$

the instructions say to first evaluate
$int_1^a e^{-xy} dy$

which comes out to the integrand of the original improper integral.

using this it seems we can rewrite the improper integral as the double integral as follows

$int{_0^infty}int{_1^a} e^{-xy}dydx$

However where I am stuck is changing the order of integration on this double integral in order to evaluate it.

This question already has an answer here:

• 
  Best way to integrate $ int_0^infty frac{e^{-at} - e^{-bt}}{t} text{d}t $
  
  2 answers
  
  

integration improper-integrals order-of-integration

integration improper-integrals order-of-integration

share|cite|improve this question

asked Jan 2 at 23:59

heironymousheironymous

41

share|cite|improve this question

asked Jan 2 at 23:59

heironymousheironymous

41

share|cite|improve this question

share|cite|improve this question

asked Jan 2 at 23:59

heironymousheironymous

41

asked Jan 2 at 23:59

heironymousheironymous

41

asked Jan 2 at 23:59

heironymousheironymous

41

41

marked as duplicate by Zacky, RRL, Lord Shark the Unknown, KReiser, Community♦ Jan 3 at 4:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Zacky, RRL, Lord Shark the Unknown, KReiser, Community♦ Jan 3 at 4:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See: en.wikipedia.org/wiki/Frullani_integral or set $a=1$ here:math.stackexchange.com/q/62254/515527
  $endgroup$
  – Zacky
  Jan 3 at 0:01
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See: en.wikipedia.org/wiki/Frullani_integral or set $a=1$ here:math.stackexchange.com/q/62254/515527
  $endgroup$
  – Zacky
  Jan 3 at 0:01
  
  
  

  
  
  
  

$begingroup$
See: en.wikipedia.org/wiki/Frullani_integral or set $a=1$ here:math.stackexchange.com/q/62254/515527
$endgroup$
– Zacky
Jan 3 at 0:01

$begingroup$
See: en.wikipedia.org/wiki/Frullani_integral or set $a=1$ here:math.stackexchange.com/q/62254/515527
$endgroup$
– Zacky
Jan 3 at 0:01

add a comment | 

1 Answer
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Let's start$$begin{align*}intlimits_0^{infty}mathrm dx,frac {e^{-x}-e^{-ax}}x & =intlimits_0^{infty}mathrm dx,intlimits_1^amathrm dy, e^{-xy}\ & =intlimits_1^amathrm dy,intlimits_0^{infty}mathrm dx, e^{-xy}end{align*}$$Letting $z=xy$ and evaluating the integral gives that$$intlimits_0^{infty}mathrm dx, e^{-xy}=frac 1y$$Hence$$intlimits_0^{infty}mathrm dx,frac {e^{-x}-e^{-ax}}xcolor{blue}{=log a}$$

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answered Jan 3 at 0:06

Frank W.Frank W.

3,3391321

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Let's start$$begin{align*}intlimits_0^{infty}mathrm dx,frac {e^{-x}-e^{-ax}}x & =intlimits_0^{infty}mathrm dx,intlimits_1^amathrm dy, e^{-xy}\ & =intlimits_1^amathrm dy,intlimits_0^{infty}mathrm dx, e^{-xy}end{align*}$$Letting $z=xy$ and evaluating the integral gives that$$intlimits_0^{infty}mathrm dx, e^{-xy}=frac 1y$$Hence$$intlimits_0^{infty}mathrm dx,frac {e^{-x}-e^{-ax}}xcolor{blue}{=log a}$$

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answered Jan 3 at 0:06

Frank W.Frank W.

3,3391321

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1

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Let's start$$begin{align*}intlimits_0^{infty}mathrm dx,frac {e^{-x}-e^{-ax}}x & =intlimits_0^{infty}mathrm dx,intlimits_1^amathrm dy, e^{-xy}\ & =intlimits_1^amathrm dy,intlimits_0^{infty}mathrm dx, e^{-xy}end{align*}$$Letting $z=xy$ and evaluating the integral gives that$$intlimits_0^{infty}mathrm dx, e^{-xy}=frac 1y$$Hence$$intlimits_0^{infty}mathrm dx,frac {e^{-x}-e^{-ax}}xcolor{blue}{=log a}$$

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answered Jan 3 at 0:06

Frank W.Frank W.

3,3391321

$endgroup$

add a comment | 

1

1

1

$begingroup$

Let's start$$begin{align*}intlimits_0^{infty}mathrm dx,frac {e^{-x}-e^{-ax}}x & =intlimits_0^{infty}mathrm dx,intlimits_1^amathrm dy, e^{-xy}\ & =intlimits_1^amathrm dy,intlimits_0^{infty}mathrm dx, e^{-xy}end{align*}$$Letting $z=xy$ and evaluating the integral gives that$$intlimits_0^{infty}mathrm dx, e^{-xy}=frac 1y$$Hence$$intlimits_0^{infty}mathrm dx,frac {e^{-x}-e^{-ax}}xcolor{blue}{=log a}$$

share|cite|improve this answer

answered Jan 3 at 0:06

Frank W.Frank W.

3,3391321

$endgroup$

Let's start$$begin{align*}intlimits_0^{infty}mathrm dx,frac {e^{-x}-e^{-ax}}x & =intlimits_0^{infty}mathrm dx,intlimits_1^amathrm dy, e^{-xy}\ & =intlimits_1^amathrm dy,intlimits_0^{infty}mathrm dx, e^{-xy}end{align*}$$Letting $z=xy$ and evaluating the integral gives that$$intlimits_0^{infty}mathrm dx, e^{-xy}=frac 1y$$Hence$$intlimits_0^{infty}mathrm dx,frac {e^{-x}-e^{-ax}}xcolor{blue}{=log a}$$

share|cite|improve this answer

answered Jan 3 at 0:06

Frank W.Frank W.

3,3391321

share|cite|improve this answer

share|cite|improve this answer

answered Jan 3 at 0:06

Frank W.Frank W.

3,3391321

answered Jan 3 at 0:06

Frank W.Frank W.

3,3391321

answered Jan 3 at 0:06

Frank W.Frank W.

3,3391321

3,3391321

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