How to compute ∫−∞∞exp⁡(−(x2−13x−1)2611x2)dx

Question

How to compute  ∫−∞∞exp⁡(−(x2−13x−1)2611x2)dx

Bernard Lacey · Accepted Answer

Let u(x)=x2−13x−1x. As x varies over R, we have  u(x) increases monotonically from −∞ at −∞ to +∞ at 0−.  u(x) increases monotonically from −∞ at 0+ to +∞ at +∞  This means as x varies, u(x) covered (−∞,∞) twice.  Let x1(u)&amp;lt;0 and x2(u)&amp;gt;0 be the two roots of the equation for a given u:  u=u(x)=x2−13x−1x⇔x2−(13+u)x−1=0  we have  x1(u)+x2(u)=13+u⇒x2−(13+u)x−1=0  we have  x1(u)+x2(u)=13+u⇒dx1du+dx2du=1  From this, we find  ∫−∞∞e−u(x)2611dx=(∫−∞0−+∫0++∞e−u(x)2611dx  =∫−∞∞e−u(x)2611(dx1du+dx2du)du  =∫−∞∞e−u(x)2611du  =611π

Jonathan Burroughs · Accepted Answer

Adding another solution owing to a friend of mine.  Through some algebra, the integral is equivalent to  ∫−∞∞exp⁡(−1611((x−x−1)−13)2)dx  Then using the following identity  ∫−∞∞f(x−x−1)dx=∫−∞∞f(x)dx  We have  ∫−∞∞exp⁡(−1611((x−x−1)−13)2)dx  =∫−∞∞exp⁡(−1611(x−13)2)dx  ∫−∞∞exp⁡(−1611x2)dx  =611π

star233 · Accepted Answer

Using identity: ∫−∞∞f(x−x−1)dx=∫−∞∞f(x)dx The problem can be generalised to evaluate ∫−∞∞exp(−(x2−bx−1)2ax2)dx=aπ Proof: It&#039;s easy to see that (x2−bx−1)2ax2=1a(x−x−1−b)2, then ∫−∞∞exp(−(x2−bx−1)2ax2)dx=∫−∞∞exp(−(x−x−1−b)2a)dx=∞−∞∞exp(−(x−b)2a)dx=∫−∞∞exp(−y2a)dy=a∫−∞∞exp(−z2)dz=aπ Therefore ∫−∞∞exp(−(x2−13x−1)2611x2)dx=611π

How to compute \int_{-\infty}^\infty exp(-\frac{(x^2-13x-1)^2}{611x^2})dx

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