Recipe

Evaluate Gamma(z) from the defining integral

Use when: you're asked for $\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}\,dt$ at a specific $z$ , using the integral itself rather than a known identity

Goal: either the integral's exact value at that $z$ , or a confirmation that it genuinely diverges there

Taught in The Gamma function

Quick recipe

Confirm $\operatorname{Re}(z)>0$ before evaluating anything
Integrate directly, or set up integration by parts
Evaluate the boundary term and any remaining limits
If Step 1 flagged divergence, confirm it directly

1
Check how the integrand behaves as $t\to0^+$ . It converges there exactly when $\operatorname{Re}(z)>0$ , and fails otherwise.

Example

At $z=0$ the integrand is $t^{-1}e^{-t}$ . It behaves like $\dfrac1t$ near $t=0$ , since $e^{-t}\to1$ there. That is the borderline exponent the $p$ -integral test already rules out.
- $\operatorname{Re}(z)>0$ — the integral converges at this $z$ . Continue to Step 2 for a genuine value.
- $\operatorname{Re}(z)\le0$ — the integral diverges at this $z$ . Skip to Step 4 and confirm the divergence directly. Do not chase a value that does not exist.
Why: this integral is undefined for $\operatorname{Re}(z)\le0$ , not just hard to compute. Checking first saves work on a value that was never there.
2
For $z=1$ , the integrand is just $e^{-t}$ , which has an elementary antiderivative. For an integer $z>1$ , use integration by parts. Take $u=t^{z-1}$ and $dv=e^{-t}\,dt$ , so $du=(z-1)t^{z-2}\,dt$ and $v=-e^{-t}$ . This strips off one power of $t$ .

Example

For $z=2$ , take $u=t$ and $dv=e^{-t}\,dt$ . Then $du=dt$ and $v=-e^{-t}$ , so $\Gamma(2)=\Bigl[-t\,e^{-t}\Bigr]_0^\infty+\int_0^\infty e^{-t}\,dt.$
- $z=1$ — no by-parts step is needed. Go straight to Step 3 and evaluate $\int_0^\infty e^{-t}\,dt$ at its limits.
- $z$ is an integer greater than $1$ — the by-parts step leaves a boundary term plus a new integral with one power of $t$ fewer. That new integral still needs Step 3's boundary check.
Why: integration by parts lowers the exponent on $t$ one step at a time. Picking $u$ and $dv$ the other way around raises the power instead.
3
Check each boundary term, and the final antiderivative, at both $t\to\infty$ and $t=0$ . Exponential decay beats any fixed power of $t$ . So a term like $t^k e^{-t}$ always vanishes as $t\to\infty$ . It is also $0$ at $t=0$ whenever $k>0$ .

Example

For $\Gamma(2)$ , the boundary term is $\bigl[-t\,e^{-t}\bigr]_0^\infty=(0)-(0)=0$ . That leaves $\Gamma(2)=\int_0^\infty e^{-t}\,dt=\Gamma(1)$ . Evaluating that piece directly gives $\bigl[-e^{-t}\bigr]_0^\infty=(0)-(-1)=1$ , so $\Gamma(2)=1$ .

Why: this is where the numeric value finally appears. Skipping the $t\to\infty$ check misses the exact place where exponential decay does the work.
4
Integrate the near- $0$ piece, using a limit as the lower bound approaches $0^+$ . Then show the result grows without bound.

Example

At $z=0$ we get $\displaystyle\int_0^1\frac{dt}t=\lim_{\varepsilon\to0^+}\Bigl[\ln t\Bigr]_\varepsilon^1=\lim_{\varepsilon\to0^+}\bigl(0-\ln\varepsilon\bigr)=\infty$ . So $\Gamma(0)$ is not defined by this integral.

Why: this turns "the integral diverges" into a checked fact instead of an assumption. It uses the same near- $t=0$ argument that Step 1 flagged as the delicate end.

Common mistakes

Trying to compute a numeric value at $z=0$ , or at a negative real part, without checking convergence first. The integral is undefined there, not just difficult.
Checking the boundary term only at $t=0$ , where it is obviously $0$ . This skips the $t\to\infty$ limit and misses the real reason it vanishes there.
Using $\Gamma(n)=(n-1)!$ as a shortcut on the integral itself before it is established. Treat it as the pattern this computation previews, not yet a proven tool.
Choosing $u$ and $dv$ backward in the by-parts step, for example $u=e^{-t}$ and $dv=t^{z-1}\,dt$ . That raises the power of $t$ instead of lowering it.