Integral of exponential of complex expression $int_-infty^inftyexpleft(-aleft[left(y+ib/2aright)^2-i^2b^2/4a^2right]right)dy$

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I have the following expression




$$
Rint_-infty^inftyexp bigg(-abigg[bigg(y+ib/2abigg)^2-i^2b^2/4a^2bigg]bigg)dy
$$
Where $R$ is real numbers and $i$ denotes complex numbers.




Which should result in the following
$$=expbigg(-b^2/4abigg)sqrtpi/a$$



I am not sure how to get to that result however, any help would be highly appreciated :)




integration complex-analysis complex-numbers




share|cite|improve this question





















  • sorry this was soled by Gauss integral as seen below :-)
    – user469216
    Aug 4 at 8:39














up vote
0
down vote

favorite












I have the following expression




$$
Rint_-infty^inftyexp bigg(-abigg[bigg(y+ib/2abigg)^2-i^2b^2/4a^2bigg]bigg)dy
$$
Where $R$ is real numbers and $i$ denotes complex numbers.




Which should result in the following
$$=expbigg(-b^2/4abigg)sqrtpi/a$$



I am not sure how to get to that result however, any help would be highly appreciated :)




integration complex-analysis complex-numbers




share|cite|improve this question





















  • sorry this was soled by Gauss integral as seen below :-)
    – user469216
    Aug 4 at 8:39












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have the following expression




$$
Rint_-infty^inftyexp bigg(-abigg[bigg(y+ib/2abigg)^2-i^2b^2/4a^2bigg]bigg)dy
$$
Where $R$ is real numbers and $i$ denotes complex numbers.




Which should result in the following
$$=expbigg(-b^2/4abigg)sqrtpi/a$$



I am not sure how to get to that result however, any help would be highly appreciated :)




integration complex-analysis complex-numbers




share|cite|improve this question













I have the following expression




$$
Rint_-infty^inftyexp bigg(-abigg[bigg(y+ib/2abigg)^2-i^2b^2/4a^2bigg]bigg)dy
$$
Where $R$ is real numbers and $i$ denotes complex numbers.




Which should result in the following
$$=expbigg(-b^2/4abigg)sqrtpi/a$$



I am not sure how to get to that result however, any help would be highly appreciated :)





integration complex-analysis complex-numbers





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 4 at 15:38









Nosrati

19.9k41644




19.9k41644









asked Jul 15 at 10:51









user469216

445




445











  • sorry this was soled by Gauss integral as seen below :-)
    – user469216
    Aug 4 at 8:39
















  • sorry this was soled by Gauss integral as seen below :-)
    – user469216
    Aug 4 at 8:39















sorry this was soled by Gauss integral as seen below :-)
– user469216
Aug 4 at 8:39




sorry this was soled by Gauss integral as seen below :-)
– user469216
Aug 4 at 8:39










1 Answer
1






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oldest

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up vote
0
down vote



accepted










Hint:
$$int_-infty^inftyexp bigg(-abigg[bigg(y+ib/2abigg)^2-i^2b^2/4a^2bigg]bigg)dy=expbigg(-b^2/4abigg) int_-infty^inftyexp bigg(-abigg[bigg(y+ib/2abigg)^2bigg]bigg)dy$$
now let $sqrtabigg(y+ib/2abigg)=u$ and after substitution use gamma function. Note that $e^-ay^2$ is even.






share|cite|improve this answer





















  • thanks good hint :-). I found another way, by splitting the function up as you did and letting $bigg(y+ib/2abigg)=u$, and then use the Gauss integral, i.e. $int_-infty^inftye^-a cdot(x+b)^2=sqrtpi/a$.
    – user469216
    Jul 15 at 20:16











  • Good point . . .
    – Nosrati
    Jul 15 at 20:54










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










Hint:
$$int_-infty^inftyexp bigg(-abigg[bigg(y+ib/2abigg)^2-i^2b^2/4a^2bigg]bigg)dy=expbigg(-b^2/4abigg) int_-infty^inftyexp bigg(-abigg[bigg(y+ib/2abigg)^2bigg]bigg)dy$$
now let $sqrtabigg(y+ib/2abigg)=u$ and after substitution use gamma function. Note that $e^-ay^2$ is even.






share|cite|improve this answer





















  • thanks good hint :-). I found another way, by splitting the function up as you did and letting $bigg(y+ib/2abigg)=u$, and then use the Gauss integral, i.e. $int_-infty^inftye^-a cdot(x+b)^2=sqrtpi/a$.
    – user469216
    Jul 15 at 20:16











  • Good point . . .
    – Nosrati
    Jul 15 at 20:54














up vote
0
down vote



accepted










Hint:
$$int_-infty^inftyexp bigg(-abigg[bigg(y+ib/2abigg)^2-i^2b^2/4a^2bigg]bigg)dy=expbigg(-b^2/4abigg) int_-infty^inftyexp bigg(-abigg[bigg(y+ib/2abigg)^2bigg]bigg)dy$$
now let $sqrtabigg(y+ib/2abigg)=u$ and after substitution use gamma function. Note that $e^-ay^2$ is even.






share|cite|improve this answer





















  • thanks good hint :-). I found another way, by splitting the function up as you did and letting $bigg(y+ib/2abigg)=u$, and then use the Gauss integral, i.e. $int_-infty^inftye^-a cdot(x+b)^2=sqrtpi/a$.
    – user469216
    Jul 15 at 20:16











  • Good point . . .
    – Nosrati
    Jul 15 at 20:54












up vote
0
down vote



accepted







up vote
0
down vote



accepted






Hint:
$$int_-infty^inftyexp bigg(-abigg[bigg(y+ib/2abigg)^2-i^2b^2/4a^2bigg]bigg)dy=expbigg(-b^2/4abigg) int_-infty^inftyexp bigg(-abigg[bigg(y+ib/2abigg)^2bigg]bigg)dy$$
now let $sqrtabigg(y+ib/2abigg)=u$ and after substitution use gamma function. Note that $e^-ay^2$ is even.






share|cite|improve this answer













Hint:
$$int_-infty^inftyexp bigg(-abigg[bigg(y+ib/2abigg)^2-i^2b^2/4a^2bigg]bigg)dy=expbigg(-b^2/4abigg) int_-infty^inftyexp bigg(-abigg[bigg(y+ib/2abigg)^2bigg]bigg)dy$$
now let $sqrtabigg(y+ib/2abigg)=u$ and after substitution use gamma function. Note that $e^-ay^2$ is even.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 15 at 11:54









Nosrati

19.9k41644




19.9k41644











  • thanks good hint :-). I found another way, by splitting the function up as you did and letting $bigg(y+ib/2abigg)=u$, and then use the Gauss integral, i.e. $int_-infty^inftye^-a cdot(x+b)^2=sqrtpi/a$.
    – user469216
    Jul 15 at 20:16











  • Good point . . .
    – Nosrati
    Jul 15 at 20:54
















  • thanks good hint :-). I found another way, by splitting the function up as you did and letting $bigg(y+ib/2abigg)=u$, and then use the Gauss integral, i.e. $int_-infty^inftye^-a cdot(x+b)^2=sqrtpi/a$.
    – user469216
    Jul 15 at 20:16











  • Good point . . .
    – Nosrati
    Jul 15 at 20:54















thanks good hint :-). I found another way, by splitting the function up as you did and letting $bigg(y+ib/2abigg)=u$, and then use the Gauss integral, i.e. $int_-infty^inftye^-a cdot(x+b)^2=sqrtpi/a$.
– user469216
Jul 15 at 20:16





thanks good hint :-). I found another way, by splitting the function up as you did and letting $bigg(y+ib/2abigg)=u$, and then use the Gauss integral, i.e. $int_-infty^inftye^-a cdot(x+b)^2=sqrtpi/a$.
– user469216
Jul 15 at 20:16













Good point . . .
– Nosrati
Jul 15 at 20:54




Good point . . .
– Nosrati
Jul 15 at 20:54












 

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