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The integral ∫0∞tp-1e-tdt=Γ'(p,x)is called an incomplete Γ'function. [Note that if x = 0, this integral isΓ'(p).] By repeated integration by parts, find several terms of the asymptotic series forΓ'(p,x).

Short Answer

Expert verified

The several terms of asymptotic series is mentioned below.Γ'(p,x)=xp-1e-x1+p-1x+p-1p-2x2+p-1p-2p-3x3+....

Step by step solution

01

Given Information

The incomplete gamma function is∫0∞tp-1e-tdt=Γ'(p,x).

02

Definition of Gamma function.

Gamma function behaves similarly to a factorial for natural numbers (a discrete set), its extension to positive real numbers (a continuous set) allows it to be used to model situations involving continuous change.

Gamma functionΓ'(p)=∫0∞xp-1e-xdx.

03

Find the terms of asymptotic series.

The incomplete gamma function is ∫0∞tp-1e-tdt=Γ'(p,x).

Partially integrate the function mentioned above.

The function becomes as follows.

Γ'(p.x)=-tp-1e-tx∞+p-1∫x∞tp-2e-tdtΓ'(p.x)=xp-1e-x+p-1∫x∞tp-2e-tdt

Integrate partially ∫x∞tp-2e-tdt.

The integration becomes as follows.

∫x∞tp-2e-tdt=tp-2e-t+p-2∫x∞tp-3e-tdt

Substitute the value of integration mentioned above in the equation given below.

Γ'(p,x)=xp-1e-x+p-1∫x∞tp-2e-tdtΓ'(p,x)=xp-1e-x+p-1-tp-2e-tx∞+p-2∫x∞tp-3e-tdtΓ'(p,x)=xp-1e-x+p-1xp-2e-x+p-1p-2∫x∞tp-3e-tdt

Integrate partially.

The integration becomes as follows.

∫x∞tp-2e-tdt=xp-3e-x+p-3∫x∞tp-4e-tdt

Substitute the value of integration mentioned above in the equation given below.

Γ'(p,x)=xp-1e-x+p-1xp-2e-x+p-1p-2∫x∞tp-3e-tdtΓ'(p,x)=xp-1e-x+p-1xp-2e-x+p-1p-2xp-3e-x+p-3∫x∞tp-4e-tdtΓ'(p,x)=xp-1e-x+p-1xp-2e-x+p-1p-2xp-3e-x+p-1p-2p-3∫x∞tp-4e-tdt

Hence, the several terms of asymptotic series is mentioned below:

Γ'(p,x)=xp-1e-x1+p-1x+p-1p-2x2+p-1p-2p-3x3+.....

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