$int_0^infty sin (a^2 x^2)dx$ picking the contour of integration

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I just have a question about what contour I should pick for this problem $$int_0^infty sin (a^2 x^2)dx$$ for positive a.



I used the freshnel integral for $int_0^infty sin (x^2)dx$ with an angle of $pi/4$. (See below)



My question is, would the angle that I would use change because of the $a^2$ addition to the problem or would I want to use the same contour and just carry the $a^2$ through the problem?



Or perhaps there is a way to take care of it right off the bat?
For example of what I mean, see this post.



Thanks for your time.



enter image description here




complex-analysis contour-integration




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    Please don't put "question" in the title. Imagine what the front page would look like if everybody did that.
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up vote
0
down vote

favorite
1












I just have a question about what contour I should pick for this problem $$int_0^infty sin (a^2 x^2)dx$$ for positive a.



I used the freshnel integral for $int_0^infty sin (x^2)dx$ with an angle of $pi/4$. (See below)



My question is, would the angle that I would use change because of the $a^2$ addition to the problem or would I want to use the same contour and just carry the $a^2$ through the problem?



Or perhaps there is a way to take care of it right off the bat?
For example of what I mean, see this post.



Thanks for your time.



enter image description here




complex-analysis contour-integration




share|cite|improve this question

















  • 2




    Please don't put "question" in the title. Imagine what the front page would look like if everybody did that.
    – joriki
    Jul 28 at 2:48












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





I just have a question about what contour I should pick for this problem $$int_0^infty sin (a^2 x^2)dx$$ for positive a.



I used the freshnel integral for $int_0^infty sin (x^2)dx$ with an angle of $pi/4$. (See below)



My question is, would the angle that I would use change because of the $a^2$ addition to the problem or would I want to use the same contour and just carry the $a^2$ through the problem?



Or perhaps there is a way to take care of it right off the bat?
For example of what I mean, see this post.



Thanks for your time.



enter image description here




complex-analysis contour-integration




share|cite|improve this question













I just have a question about what contour I should pick for this problem $$int_0^infty sin (a^2 x^2)dx$$ for positive a.



I used the freshnel integral for $int_0^infty sin (x^2)dx$ with an angle of $pi/4$. (See below)



My question is, would the angle that I would use change because of the $a^2$ addition to the problem or would I want to use the same contour and just carry the $a^2$ through the problem?



Or perhaps there is a way to take care of it right off the bat?
For example of what I mean, see this post.



Thanks for your time.



enter image description here





complex-analysis contour-integration





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share|cite|improve this question




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edited Jul 28 at 3:04
























asked Jul 28 at 2:22









MathIsHard

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  • 2




    Please don't put "question" in the title. Imagine what the front page would look like if everybody did that.
    – joriki
    Jul 28 at 2:48












  • 2




    Please don't put "question" in the title. Imagine what the front page would look like if everybody did that.
    – joriki
    Jul 28 at 2:48







2




2




Please don't put "question" in the title. Imagine what the front page would look like if everybody did that.
– joriki
Jul 28 at 2:48




Please don't put "question" in the title. Imagine what the front page would look like if everybody did that.
– joriki
Jul 28 at 2:48










1 Answer
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Substitute $u=ax$:



$$
int_0^inftysinleft(a^2x^2right)mathrm dx=frac1aint_0^inftysinleft(a^2x^2right)mathrm d(ax)=frac1aint_0^inftysin u^2mathrm du;.
$$






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  • You're awesome. Thank you :)
    – MathIsHard
    Jul 28 at 5:24










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










Substitute $u=ax$:



$$
int_0^inftysinleft(a^2x^2right)mathrm dx=frac1aint_0^inftysinleft(a^2x^2right)mathrm d(ax)=frac1aint_0^inftysin u^2mathrm du;.
$$






share|cite|improve this answer





















  • You're awesome. Thank you :)
    – MathIsHard
    Jul 28 at 5:24














up vote
3
down vote



accepted










Substitute $u=ax$:



$$
int_0^inftysinleft(a^2x^2right)mathrm dx=frac1aint_0^inftysinleft(a^2x^2right)mathrm d(ax)=frac1aint_0^inftysin u^2mathrm du;.
$$






share|cite|improve this answer





















  • You're awesome. Thank you :)
    – MathIsHard
    Jul 28 at 5:24












up vote
3
down vote



accepted







up vote
3
down vote



accepted






Substitute $u=ax$:



$$
int_0^inftysinleft(a^2x^2right)mathrm dx=frac1aint_0^inftysinleft(a^2x^2right)mathrm d(ax)=frac1aint_0^inftysin u^2mathrm du;.
$$






share|cite|improve this answer













Substitute $u=ax$:



$$
int_0^inftysinleft(a^2x^2right)mathrm dx=frac1aint_0^inftysinleft(a^2x^2right)mathrm d(ax)=frac1aint_0^inftysin u^2mathrm du;.
$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 28 at 2:51









joriki

164k10179328




164k10179328











  • You're awesome. Thank you :)
    – MathIsHard
    Jul 28 at 5:24
















  • You're awesome. Thank you :)
    – MathIsHard
    Jul 28 at 5:24















You're awesome. Thank you :)
– MathIsHard
Jul 28 at 5:24




You're awesome. Thank you :)
– MathIsHard
Jul 28 at 5:24












 

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