Exponential integral with square root [closed]

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Can this be solved?



$$Y=int_0^infty Ae^-Bsqrttmathrmdt$$
Where $A$ and $B$ are constants.




integration definite-integrals improper-integrals




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closed as off-topic by Theoretical Economist, Jendrik Stelzner, Brahadeesh, Sil, amWhy Aug 19 at 10:23


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  • Please, use MathJax (i.e. LaTeX commands) for mathematical notations.
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    Aug 14 at 8:24














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Can this be solved?



$$Y=int_0^infty Ae^-Bsqrttmathrmdt$$
Where $A$ and $B$ are constants.




integration definite-integrals improper-integrals




share|cite|improve this question














closed as off-topic by Theoretical Economist, Jendrik Stelzner, Brahadeesh, Sil, amWhy Aug 19 at 10:23


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Theoretical Economist, Jendrik Stelzner, Brahadeesh, Sil, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Please, use MathJax (i.e. LaTeX commands) for mathematical notations.
    – Taroccoesbrocco
    Aug 14 at 8:24












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Can this be solved?



$$Y=int_0^infty Ae^-Bsqrttmathrmdt$$
Where $A$ and $B$ are constants.




integration definite-integrals improper-integrals




share|cite|improve this question














Can this be solved?



$$Y=int_0^infty Ae^-Bsqrttmathrmdt$$
Where $A$ and $B$ are constants.





integration definite-integrals improper-integrals





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 14 at 8:58









Davide Morgante

2,282322




2,282322










asked Aug 14 at 8:20









Ernest Ong

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31




closed as off-topic by Theoretical Economist, Jendrik Stelzner, Brahadeesh, Sil, amWhy Aug 19 at 10:23


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Theoretical Economist, Jendrik Stelzner, Brahadeesh, Sil, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Theoretical Economist, Jendrik Stelzner, Brahadeesh, Sil, amWhy Aug 19 at 10:23


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Theoretical Economist, Jendrik Stelzner, Brahadeesh, Sil, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.











  • Please, use MathJax (i.e. LaTeX commands) for mathematical notations.
    – Taroccoesbrocco
    Aug 14 at 8:24
















  • Please, use MathJax (i.e. LaTeX commands) for mathematical notations.
    – Taroccoesbrocco
    Aug 14 at 8:24















Please, use MathJax (i.e. LaTeX commands) for mathematical notations.
– Taroccoesbrocco
Aug 14 at 8:24




Please, use MathJax (i.e. LaTeX commands) for mathematical notations.
– Taroccoesbrocco
Aug 14 at 8:24










1 Answer
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up vote
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Let us do a change of variable: choose $x=-Bsqrtt$ then we have $$t=fracx^2B^2\ mathrmdt = frac2xB^2mathrmdx$$ Substituting in the integral $$int_0^infty Ae^Bsqrttmathrmdt = frac2AB^2int_0^infty xe^-xmathrmdx = frac2AB^2int_0^infty x^1-1e^-xmathrmdx = frac2AB^2Gamma(1)=frac2AB^2$$
where $Gamma(z)$ is the Euler Gamma function, defined as $$Gamma(z)= int_0^infty t^z-1e^-tmathrmdt$$



We could solve the last integral without the Euler gamma function by using integration by parts, mainly $$int_0^infty underbracex_funderbracee^-x_g'mathrmdx = -underbrace_0^infty_textis zero+int_0^infty e^-xmathrmdx = -e^-x|_0^infty = 1$$ Remember that integration by parts says $$int f(x)g'(x)mathrmdx = f(x)g(x) - int f'(x)g(x)mathrmdx$$






share|cite|improve this answer




















  • Thank you so much, Davide Morgante! It is a very detailed explanation. Highly appreciate it!
    – Ernest Ong
    Aug 15 at 0:39










  • You're welcome!
    – Davide Morgante
    Aug 15 at 7:47

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Let us do a change of variable: choose $x=-Bsqrtt$ then we have $$t=fracx^2B^2\ mathrmdt = frac2xB^2mathrmdx$$ Substituting in the integral $$int_0^infty Ae^Bsqrttmathrmdt = frac2AB^2int_0^infty xe^-xmathrmdx = frac2AB^2int_0^infty x^1-1e^-xmathrmdx = frac2AB^2Gamma(1)=frac2AB^2$$
where $Gamma(z)$ is the Euler Gamma function, defined as $$Gamma(z)= int_0^infty t^z-1e^-tmathrmdt$$



We could solve the last integral without the Euler gamma function by using integration by parts, mainly $$int_0^infty underbracex_funderbracee^-x_g'mathrmdx = -underbrace_0^infty_textis zero+int_0^infty e^-xmathrmdx = -e^-x|_0^infty = 1$$ Remember that integration by parts says $$int f(x)g'(x)mathrmdx = f(x)g(x) - int f'(x)g(x)mathrmdx$$






share|cite|improve this answer




















  • Thank you so much, Davide Morgante! It is a very detailed explanation. Highly appreciate it!
    – Ernest Ong
    Aug 15 at 0:39










  • You're welcome!
    – Davide Morgante
    Aug 15 at 7:47














up vote
1
down vote



accepted










Let us do a change of variable: choose $x=-Bsqrtt$ then we have $$t=fracx^2B^2\ mathrmdt = frac2xB^2mathrmdx$$ Substituting in the integral $$int_0^infty Ae^Bsqrttmathrmdt = frac2AB^2int_0^infty xe^-xmathrmdx = frac2AB^2int_0^infty x^1-1e^-xmathrmdx = frac2AB^2Gamma(1)=frac2AB^2$$
where $Gamma(z)$ is the Euler Gamma function, defined as $$Gamma(z)= int_0^infty t^z-1e^-tmathrmdt$$



We could solve the last integral without the Euler gamma function by using integration by parts, mainly $$int_0^infty underbracex_funderbracee^-x_g'mathrmdx = -underbrace_0^infty_textis zero+int_0^infty e^-xmathrmdx = -e^-x|_0^infty = 1$$ Remember that integration by parts says $$int f(x)g'(x)mathrmdx = f(x)g(x) - int f'(x)g(x)mathrmdx$$






share|cite|improve this answer




















  • Thank you so much, Davide Morgante! It is a very detailed explanation. Highly appreciate it!
    – Ernest Ong
    Aug 15 at 0:39










  • You're welcome!
    – Davide Morgante
    Aug 15 at 7:47












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Let us do a change of variable: choose $x=-Bsqrtt$ then we have $$t=fracx^2B^2\ mathrmdt = frac2xB^2mathrmdx$$ Substituting in the integral $$int_0^infty Ae^Bsqrttmathrmdt = frac2AB^2int_0^infty xe^-xmathrmdx = frac2AB^2int_0^infty x^1-1e^-xmathrmdx = frac2AB^2Gamma(1)=frac2AB^2$$
where $Gamma(z)$ is the Euler Gamma function, defined as $$Gamma(z)= int_0^infty t^z-1e^-tmathrmdt$$



We could solve the last integral without the Euler gamma function by using integration by parts, mainly $$int_0^infty underbracex_funderbracee^-x_g'mathrmdx = -underbrace_0^infty_textis zero+int_0^infty e^-xmathrmdx = -e^-x|_0^infty = 1$$ Remember that integration by parts says $$int f(x)g'(x)mathrmdx = f(x)g(x) - int f'(x)g(x)mathrmdx$$






share|cite|improve this answer












Let us do a change of variable: choose $x=-Bsqrtt$ then we have $$t=fracx^2B^2\ mathrmdt = frac2xB^2mathrmdx$$ Substituting in the integral $$int_0^infty Ae^Bsqrttmathrmdt = frac2AB^2int_0^infty xe^-xmathrmdx = frac2AB^2int_0^infty x^1-1e^-xmathrmdx = frac2AB^2Gamma(1)=frac2AB^2$$
where $Gamma(z)$ is the Euler Gamma function, defined as $$Gamma(z)= int_0^infty t^z-1e^-tmathrmdt$$



We could solve the last integral without the Euler gamma function by using integration by parts, mainly $$int_0^infty underbracex_funderbracee^-x_g'mathrmdx = -underbrace_0^infty_textis zero+int_0^infty e^-xmathrmdx = -e^-x|_0^infty = 1$$ Remember that integration by parts says $$int f(x)g'(x)mathrmdx = f(x)g(x) - int f'(x)g(x)mathrmdx$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 14 at 8:39









Davide Morgante

2,282322




2,282322











  • Thank you so much, Davide Morgante! It is a very detailed explanation. Highly appreciate it!
    – Ernest Ong
    Aug 15 at 0:39










  • You're welcome!
    – Davide Morgante
    Aug 15 at 7:47
















  • Thank you so much, Davide Morgante! It is a very detailed explanation. Highly appreciate it!
    – Ernest Ong
    Aug 15 at 0:39










  • You're welcome!
    – Davide Morgante
    Aug 15 at 7:47















Thank you so much, Davide Morgante! It is a very detailed explanation. Highly appreciate it!
– Ernest Ong
Aug 15 at 0:39




Thank you so much, Davide Morgante! It is a very detailed explanation. Highly appreciate it!
– Ernest Ong
Aug 15 at 0:39












You're welcome!
– Davide Morgante
Aug 15 at 7:47




You're welcome!
– Davide Morgante
Aug 15 at 7:47


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