Question

integration of e to the power 2 x minus e to the power minus 2 X upon e to the power 2 X + e to the power minus 2 x​

Answer 1

Answer:

Step-by-step explanation:

Let me tell you one of my (maybe absurd) feeling for this problem. Let me show you one of the formal derivation of the value of

∫−∞∞e−x2dx=π−−√

Substituting, x = y

Let,

I=∫−∞∞e−x2dx=∫−∞∞e−y2dy

Thus,

I2=I.I=∫−∞∞e−x2dx∫−∞∞e−y2dy

⟹I2=∫−∞∞∫−∞∞e−(x2+y2)dxdy

Substituting, x2+y2=r2

Thus,

dxdy=Jdrdθ=rdrdθ

J is a mathematical entity known as Jacobian. If you haven't heard of it, it's nothing to worry about; just keep that in mind this is sort of a “Transformation coefficient” i.e., a factor that is necessary to change the coordinates from (x,y) to (u,v). Since, u = r and v = θ, so J = r.

⟹I2=∫r=0∞∫θ=02πe−r2rdrdθ

⟹I2=∫r=0∞re−r2dr∫θ=02πdθ

⟹I2=−e−r22∣∣∣∞0.θ|2π0

⟹I2=12.2π=π

Thus,

I=π−−√

Or,

∫−∞∞e−x2dx=π−−√

I hope following these steps would answer your question completely. But let me share one of my (don't know what but) weird ideas about solving this integral